Before we start we note that $\mathcal{O}_{X, x}$ is a regular local ring for all $x \in D$. This follows from Algebra, Lemma \ref{algebra-lemma-regular-mod-x} and our assumption that $\mathcal{O}_{D, x}$ is regular. Below we will also use that regular rings are normal, see Algebra, Lemma \ref{algebra-lemma-regular-normal}. \medskip\noindent To prove the lemma we may work locally on $X$. Thus we may assume $X = \Spec(A)$ and $D \subset X$ is given by a nonzerodivisor $f \in A$. Then $Y = \Spec(B)$ as a finite \'etale scheme over $A_f$. Let $\mathfrak p_1, \ldots, \mathfrak p_r$ be the minimal primes of $A$ over $f$. Then $A_i = A_{\mathfrak p_i}$ is a discrete valuation ring; denote its fraction field $K_i$. By assumption $$ K_i \otimes_{A_f} B = \prod L_{ij} $$ is a finite product of fields each tamely ramified with respect to $A_i$. Choose $e \geq 1$ sufficiently divisible (namely, divisible by all ramification indices for $L_{ij}$ over $A_i$ as in More on Algebra, Remark \ref{more-algebra-remark-finite-separable-extension}). Warning: at this point we do not know that $e$ is invertible on $A$. \medskip\noindent Consider the finite free $A$-algebra $$ A' = A[x]/(x^e - f) $$ Observe that $f' = x$ is a nonzerodivisor in $A'$ and that $A'/f'A' \cong A/fA$ is a regular ring. Set $B' = B \otimes_A A' = B \otimes_{A_f} A'_{f'}$. By Abhyankar's lemma (More on Algebra, Lemma \ref{more-algebra-lemma-abhyankar}) we see that $\Spec(B')$ is unramified over $\Spec(A')$ in codimension $1$. Namely, by Lemma \ref{lemma-pullback-tame-codim1} we see that $\Spec(B')$ is still at least tamely ramified over $\Spec(A')$ in codimension $1$. But Abhyankar's lemma tells us that the ramification indices have all become equal to $1$. By Lemma \ref{lemma-purity-one-divisor} we conclude that $\Spec(B') \to \Spec(A'_{f'})$ extends to a finite \'etale morphism $\Spec(C) \to \Spec(A')$. \medskip\noindent For a point $x \in D$ corresponding to $\mathfrak p \in V(f)$ denote $A^{sh}$ a strict henselization of $A_\mathfrak p = \mathcal{O}_{X, x}$. Observe that $A^{sh}$ and $A^{sh}/fA^{sh} = (A/fA)^{sh}$ (Algebra, Lemma \ref{algebra-lemma-quotient-strict-henselization}) are regular local rings, see More on Algebra, Lemma \ref{more-algebra-lemma-henselization-regular}. Observe that $A'$ has a unique prime $\mathfrak p'$ lying over $\mathfrak p$ with identical residue field. Thus $$ (A')^{sh} = A^{sh} \otimes_A A' = A^{sh}[x]/(x^e - f) $$ is a strictly henselian local ring finite over $A^{sh}$ (Algebra, Lemma \ref{algebra-lemma-quasi-finite-strict-henselization}). Since $f'$ is a nonzerodivisor in $(A')^{sh}$ and since $(A')^{sh}/f'(A')^{sh} = A^{sh}/fA^{sh}$ is regular, we conclude that $(A')^{sh}$ is a regular local ring (see above). Observe that the induced extension $$ Q(A^{sh}) \subset Q((A')^{sh}) = Q(A^{sh})[x]/(x^e - f) $$ of fraction fields has degree $e$ (and not less). Since $A' \to C$ is finite \'etale we see that $A^{sh} \otimes_A C$ is a finite product of copies of $(A')^{sh}$ (Algebra, Lemma \ref{algebra-lemma-mop-up-strictly-henselian}). We have the inclusions $$ A^{sh}_f \subset A^{sh} \otimes_A B \subset A^{sh} \otimes_A B' = A^{sh} \otimes_A C_{f'} $$ and each of these rings is Noetherian and normal; this follows from Algebra, Lemma \ref{algebra-lemma-normal-goes-up} for the ring in the middle. Taking total quotient rings, using the product decomposition of $A^{sh} \otimes_A C$ and using Fields, Lemma \ref{fields-lemma-subfields-kummer} we conclude that there is an isomorphism $$ Q(A^{sh}) \otimes_A B \cong \prod\nolimits_{i \in I} F_i,\quad F_i \cong Q(A^{sh})[x]/(x^{e_i} - f) $$ of $Q(A^{sh})$-algebras for some finite set $I$ and integers $e_i | e$. Since $A^{sh} \otimes_A B$ is a normal ring, it must be the integral closure of $A^{sh}$ in its total quotient ring. We conclude that we have an isomorphism $$ A^{sh} \otimes_A B \cong \prod A^{sh}_f[x]/(x^{e_i} - f) $$ over $A^{sh}_f$ because the algebras $A^{sh}[x]/(x^{e_i} - f)$ are regular and hence normal. The discriminant of $A^{sh}[x]/(x^{e_i} - f)$ over $A^{sh}$ is $e_i^{e_i}f^{e_i - 1}$ (up to sign; calculation omitted). Since $A_f \to B$ is finite \'etale we see that $e_i$ must be invertible in $A^{sh}_f$. On the other hand, since $A_f \to B$ is tamely ramified over $\Spec(A)$ in codimension $1$, by Lemma \ref{lemma-pullback-tame-codim1} the ring map $A^{sh}_f \to A^{sh} \otimes_A B$ is tamely ramified over $\Spec(A^{sh})$ in codimension $1$. This implies $e_i$ is nonzero in $A^{sh}/fA^{sh}$ (as it must map to an invertible element of the fraction field of this domain by definition of tamely ramified extensions). We conclude that $V(e_i) \subset \Spec(A^{sh})$ has codimension $\geq 2$ which is absurd unless it is empty. In other words, $e_i$ is an invertible element of $A^{sh}$. We conclude that the pullback of $Y$ to $\Spec(A^{sh})$ is indeed a finite disjoint union of standard tamely ramified morphisms. \medskip\noindent To finish the proof, we write $A^{sh} = \colim A_\lambda$ as a filtered colimit of \'etale $A$-algebras $A_\lambda$. The isomorphism $$ A^{sh} \otimes_A B \cong \prod\nolimits_{i \in I} A^{sh}_f[x]/(x^{e_i} - f) $$ descends to an isomorphism $$ A_\lambda \otimes_A B \cong \prod\nolimits_{i \in I} (A_\lambda)_f[x]/(x^{e_i} - f) $$ for suitably large $\lambda$. After increasing $\lambda$ a bit more we may assume $e_i$ is invertible in $A_\lambda$. Then $\Spec(A_\lambda) \to \Spec(A)$ is the desired \'etale neighbourhood of $x$ and the proof is complete.