The following proof is a little roundabout and there may be ways to shorten it. \medskip\noindent Step 1. By Definition \ref{definition-etale} there exists a $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is \'etale. Thus we may assume that $S$ is \'etale over $R$. \medskip\noindent Step 2. By Lemma \ref{lemma-etale} there exists an \'etale ring map $R_0 \to S_0$ with $R_0$ of finite type over $\mathbf{Z}$, and a ring map $R_0 \to R$ such that $R = R \otimes_{R_0} S_0$. Denote $\mathfrak q_0$ the prime of $S_0$ corresponding to $\mathfrak q$. If we show the result for $(R_0 \to S_0, \mathfrak q_0)$ then the result follows for $(R \to S, \mathfrak q)$ by base change. Hence we may assume that $R$ is Noetherian. \medskip\noindent Step 3. Note that $R \to S$ is quasi-finite by Lemma \ref{lemma-etale-quasi-finite}. By Lemma \ref{lemma-quasi-finite-open-integral-closure} there exists a finite ring map $R \to S'$, an $R$-algebra map $S' \to S$, an element $g' \in S'$ such that $g' \not \in \mathfrak q$ such that $S' \to S$ induces an isomorphism $S'_{g'} \cong S_{g'}$. (Note that of course $S'$ is not \'etale over $R$ in general.) Thus we may assume that (a) $R$ is Noetherian, (b) $R \to S$ is finite and (c) $R \to S$ is \'etale at $\mathfrak q$ (but no longer necessarily \'etale at all primes). \medskip\noindent Step 4. Let $\mathfrak p \subset R$ be the prime corresponding to $\mathfrak q$. Consider the fibre ring $S \otimes_R \kappa(\mathfrak p)$. This is a finite algebra over $\kappa(\mathfrak p)$. Hence it is Artinian (see Lemma \ref{lemma-finite-dimensional-algebra}) and so a finite product of local rings $$ S \otimes_R \kappa(\mathfrak p) = \prod\nolimits_{i = 1}^n A_i $$ see Proposition \ref{proposition-dimension-zero-ring}. One of the factors, say $A_1$, is the local ring $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ which is isomorphic to $\kappa(\mathfrak q)$, see Lemma \ref{lemma-etale-at-prime}. The other factors correspond to the other primes, say $\mathfrak q_2, \ldots, \mathfrak q_n$ of $S$ lying over $\mathfrak p$. \medskip\noindent Step 5. We may choose a nonzero element $\alpha \in \kappa(\mathfrak q)$ which generates the finite separable field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$ (so even if the field extension is trivial we do not allow $\alpha = 0$). Note that for any $\lambda \in \kappa(\mathfrak p)^*$ the element $\lambda \alpha$ also generates $\kappa(\mathfrak q)$ over $\kappa(\mathfrak p)$. Consider the element $$ \overline{t} = (\alpha, 0, \ldots, 0) \in \prod\nolimits_{i = 1}^n A_i = S \otimes_R \kappa(\mathfrak p). $$ After possibly replacing $\alpha$ by $\lambda \alpha$ as above we may assume that $\overline{t}$ is the image of $t \in S$. Let $I \subset R[x]$ be the kernel of the $R$-algebra map $R[x] \to S$ which maps $x$ to $t$. Set $S' = R[x]/I$, so $S' \subset S$. Here is a diagram $$ \xymatrix{ R[x] \ar[r] & S' \ar[r] & S \\ R \ar[u] \ar[ru] \ar[rru] & & } $$ By construction the primes $\mathfrak q_j$, $j \geq 2$ of $S$ all lie over the prime $(\mathfrak p, x)$ of $R[x]$, whereas the prime $\mathfrak q$ lies over a different prime of $R[x]$ because $\alpha \not = 0$. \medskip\noindent Step 6. Denote $\mathfrak q' \subset S'$ the prime of $S'$ corresponding to $\mathfrak q$. By the above $\mathfrak q$ is the only prime of $S$ lying over $\mathfrak q'$. Thus we see that $S_{\mathfrak q} = S_{\mathfrak q'}$, see Lemma \ref{lemma-unique-prime-over-localize-below} (we have going up for $S' \to S$ by Lemma \ref{lemma-integral-going-up} since $S' \to S$ is finite as $R \to S$ is finite). It follows that $S'_{\mathfrak q'} \to S_{\mathfrak q}$ is finite and injective as the localization of the finite injective ring map $S' \to S$. Consider the maps of local rings $$ R_{\mathfrak p} \to S'_{\mathfrak q'} \to S_{\mathfrak q} $$ The second map is finite and injective. We have $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} = \kappa(\mathfrak q)$, see Lemma \ref{lemma-etale-at-prime}. Hence a fortiori $S_{\mathfrak q}/\mathfrak q'S_{\mathfrak q} = \kappa(\mathfrak q)$. Since $$ \kappa(\mathfrak p) \subset \kappa(\mathfrak q') \subset \kappa(\mathfrak q) $$ and since $\alpha$ is in the image of $\kappa(\mathfrak q')$ in $\kappa(\mathfrak q)$ we conclude that $\kappa(\mathfrak q') = \kappa(\mathfrak q)$. Hence by Nakayama's Lemma \ref{lemma-NAK} applied to the $S'_{\mathfrak q'}$-module map $S'_{\mathfrak q'} \to S_{\mathfrak q}$, the map $S'_{\mathfrak q'} \to S_{\mathfrak q}$ is surjective. In other words, $S'_{\mathfrak q'} \cong S_{\mathfrak q}$. \medskip\noindent Step 7. By Lemma \ref{lemma-isomorphic-local-rings} there exist $g \in S$, $g \not \in \mathfrak q$ and $g' \in S'$, $g' \not \in \mathfrak q'$ such that $S'_{g'} \cong S_g$. As $R$ is Noetherian the ring $S'$ is finite over $R$ because it is an $R$-submodule of the finite $R$-module $S$. Hence after replacing $S$ by $S'$ we may assume that (a) $R$ is Noetherian, (b) $S$ finite over $R$, (c) $S$ is \'etale over $R$ at $\mathfrak q$, and (d) $S = R[x]/I$. \medskip\noindent Step 8. Consider the ring $S \otimes_R \kappa(\mathfrak p) = \kappa(\mathfrak p)[x]/\overline{I}$ where $\overline{I} = I \cdot \kappa(\mathfrak p)[x]$ is the ideal generated by $I$ in $\kappa(\mathfrak p)[x]$. As $\kappa(\mathfrak p)[x]$ is a PID we know that $\overline{I} = (\overline{h})$ for some monic $\overline{h} \in \kappa(\mathfrak p)[x]$. After replacing $\overline{h}$ by $\lambda \cdot \overline{h}$ for some $\lambda \in \kappa(\mathfrak p)$ we may assume that $\overline{h}$ is the image of some $h \in I \subset R[x]$. (The problem is that we do not know if we may choose $h$ monic.) Also, as in Step 4 we know that $S \otimes_R \kappa(\mathfrak p) = A_1 \times \ldots \times A_n$ with $A_1 = \kappa(\mathfrak q)$ a finite separable extension of $\kappa(\mathfrak p)$ and $A_2, \ldots, A_n$ local. This implies that $$ \overline{h} = \overline{h}_1 \overline{h}_2^{e_2} \ldots \overline{h}_n^{e_n} $$ for certain pairwise coprime irreducible monic polynomials $\overline{h}_i \in \kappa(\mathfrak p)[x]$ and certain $e_2, \ldots, e_n \geq 1$. Here the numbering is chosen so that $A_i = \kappa(\mathfrak p)[x]/(\overline{h}_i^{e_i})$ as $\kappa(\mathfrak p)[x]$-algebras. Note that $\overline{h}_1$ is the minimal polynomial of $\alpha \in \kappa(\mathfrak q)$ and hence is a separable polynomial (its derivative is prime to itself). \medskip\noindent Step 9. Let $m \in I$ be a monic element; such an element exists because the ring extension $R \to R[x]/I$ is finite hence integral. Denote $\overline{m}$ the image in $\kappa(\mathfrak p)[x]$. We may factor $$ \overline{m} = \overline{k} \overline{h}_1^{d_1} \overline{h}_2^{d_2} \ldots \overline{h}_n^{d_n} $$ for some $d_1 \geq 1$, $d_j \geq e_j$, $j = 2, \ldots, n$ and $\overline{k} \in \kappa(\mathfrak p)[x]$ prime to all the $\overline{h}_i$. Set $f = m^l + h$ where $l \deg(m) > \deg(h)$, and $l \geq 2$. Then $f$ is monic as a polynomial over $R$. Also, the image $\overline{f}$ of $f$ in $\kappa(\mathfrak p)[x]$ factors as $$ \overline{f} = \overline{h}_1 \overline{h}_2^{e_2} \ldots \overline{h}_n^{e_n} + \overline{k}^l \overline{h}_1^{ld_1} \overline{h}_2^{ld_2} \ldots \overline{h}_n^{ld_n} = \overline{h}_1(\overline{h}_2^{e_2} \ldots \overline{h}_n^{e_n} + \overline{k}^l \overline{h}_1^{ld_1 - 1} \overline{h}_2^{ld_2} \ldots \overline{h}_n^{ld_n}) = \overline{h}_1 \overline{w} $$ with $\overline{w}$ a polynomial relatively prime to $\overline{h}_1$. Set $g = f'$ (the derivative with respect to $x$). \medskip\noindent Step 10. The ring map $R[x] \to S = R[x]/I$ has the properties: (1) it maps $f$ to zero, and (2) it maps $g$ to an element of $S \setminus \mathfrak q$. The first assertion is clear since $f$ is an element of $I$. For the second assertion we just have to show that $g$ does not map to zero in $\kappa(\mathfrak q) = \kappa(\mathfrak p)[x]/(\overline{h}_1)$. The image of $g$ in $\kappa(\mathfrak p)[x]$ is the derivative of $\overline{f}$. Thus (2) is clear because $$ \overline{g} = \frac{\text{d}\overline{f}}{\text{d}x} = \overline{w}\frac{\text{d}\overline{h}_1}{\text{d}x} + \overline{h}_1\frac{\text{d}\overline{w}}{\text{d}x}, $$ $\overline{w}$ is prime to $\overline{h}_1$ and $\overline{h}_1$ is separable. \medskip\noindent Step 11. We conclude that $\varphi : R[x]/(f) \to S$ is a surjective ring map, $R[x]_g/(f)$ is \'etale over $R$ (because it is standard \'etale, see Lemma \ref{lemma-standard-etale}) and $\varphi(g) \not \in \mathfrak q$. Pick an element $g' \in R[x]/(f)$ such that also $\varphi(g') \not \in \mathfrak q$ and $S_{\varphi(g')}$ is \'etale over $R$ (which exists since $S$ is \'etale over $R$ at $\mathfrak q$). Then the ring map $R[x]_{gg'}/(f) \to S_{\varphi(gg')}$ is a surjective map of \'etale algebras over $R$. Hence it is \'etale by Lemma \ref{lemma-map-between-etale}. Hence it is a localization by Lemma \ref{lemma-surjective-flat-finitely-presented}. Thus a localization of $S$ at an element not in $\mathfrak q$ is isomorphic to a localization of a standard \'etale algebra over $R$ which is what we wanted to show.