By Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-points-open} the subset $U \subset X$ of points where $f$ is quasi-finite is open. The lemma is equivalent to \begin{enumerate} \item[(a)] $U' = f'(U) \subset S'$ is open, \item[(b)] $U = f^{-1}(U')$, and \item[(c)] $U \to U'$ is an isomorphism. \end{enumerate} Let $x \in U$ be arbitrary. We claim there exists an open neighbourhood $f'(x) \in V \subset S'$ such that $(f')^{-1}V \to V$ is an isomorphism. We first prove the claim implies the lemma. Namely, then $(f')^{-1}V \cong V$ is both locally of finite type over $S$ (as an open subscheme of $X$) and for $v \in V$ the residue field extension $\kappa(v) \supset \kappa(\nu(v))$ is algebraic (as $V \subset S'$ and $S'$ is integral over $S$). Hence the fibres of $V \to S$ are discrete (Morphisms, Lemma \ref{morphisms-lemma-algebraic-residue-field-extension-closed-point-fibre}) and $(f')^{-1}V \to S$ is locally quasi-finite (Morphisms, Lemma \ref{morphisms-lemma-locally-quasi-finite-fibres}). This implies $(f')^{-1}V \subset U$ and $V \subset U'$. Since $x$ was arbitrary we see that (a), (b), and (c) are true. \medskip\noindent Let $s = f(x)$. Let $(T, t) \to (S, s)$ be an elementary \'etale neighbourhood. Denote by a subscript ${}_T$ the base change to $T$. Let $y = (x, t) \in X_T$ be the unique point in the fibre $X_t$ lying over $x$. Note that $U_T \subset X_T$ is the set of points where $f_T$ is quasi-finite, see Morphisms, Lemma \ref{morphisms-lemma-base-change-quasi-finite}. Note that $$ X_T \xrightarrow{f'_T} S'_T \xrightarrow{\nu_T} T $$ is the normalization of $T$ in $X_T$, see Lemma \ref{lemma-normalization-smooth-localization}. Suppose that the claim holds for $y \in U_T \subset X_T \to S'_T \to T$, i.e., suppose that we can find an open neighbourhood $f'_T(y) \in V' \subset S'_T$ such that $(f'_T)^{-1}V' \to V'$ is an isomorphism. The morphism $S'_T \to S'$ is \'etale hence the image $V \subset S'$ of $V'$ is open. Observe that $f'(x) \in V$ as $f'_T(y) \in V'$. Observe that $$ \xymatrix{ (f'_T)^{-1}V' \ar[r] \ar[d] & (f')^{-1}(V) \ar[d] \\ V' \ar[r] & V } $$ is a fibre square (as $S'_T \times_{S'} X = X_T$). Since the left vertical arrow is an isomorphism and $\{V' \to V\}$ is a \'etale covering, we conclude that the right vertical arrow is an isomorphism by Descent, Lemma \ref{descent-lemma-descending-property-isomorphism}. In other words, the claim holds for $x \in U \subset X \to S' \to S$. \medskip\noindent By the result of the previous paragraph we may replace $S$ by an elementary \'etale neighbourhood of $s = f(x)$ in order to prove the claim. Thus we may assume there is a decomposition $$ X = V \amalg W $$ into open and closed subschemes where $V \to S$ is finite and $x \in V$, see Lemma \ref{lemma-etale-splits-off-quasi-finite-part-technical}. Since $X$ is a disjoint union of $V$ and $W$ over $S$ and since $V \to S$ is finite we see that the normalization of $S$ in $X$ is the morphism $$ X = V \amalg W \longrightarrow V \amalg W' \longrightarrow S $$ where $W'$ is the normalization of $S$ in $W$, see Morphisms, Lemmas \ref{morphisms-lemma-normalization-in-disjoint-union}, \ref{morphisms-lemma-finite-integral}, and \ref{morphisms-lemma-normalization-in-integral}. The claim follows and we win.