Let $\pi \in A$ be a uniformizer. Let $\pi_1$ be a uniformizer of $(A_1)_{\mathfrak m_i}$. Write $\pi = u \pi_1^{e_1}$ with $u$ a unit of $(A_1)_{\mathfrak m_i}$ and $e_1$ the ramification index of $A \subset (A_1)_{\mathfrak m_i}$. \medskip\noindent Claim: we may assume that $u$ is an $e$th power in $K_1$. Namely, let $K_2$ be an extension of $K_1$ obtained by adjoining a root of $x^e = u$; thus $K_2$ is a factor of $K_1[x]/(x^e - u)$. Then $K_2/K_1$ is a finite separable extension (by our assumption on $e$) and hence $A_1 \subset A_2$ is finite. Since $(A_1)_{\mathfrak m_i} \to (A_1)_{\mathfrak m_i}[x]/(x^e - u)$ is finite \'etale (as $e$ is prime to the residue characteristic and $u$ a unit) we conclude that $(A_2)_{\mathfrak m_i}$ is a factor of a finite \'etale extension of $(A_1)_{\mathfrak m_i}$ hence finite \'etale over $(A_1)_{\mathfrak m_i}$ itself. The same reasoning shows that $B_1 \subset B_2$ induces finite \'etale extensions $(B_1)_{\mathfrak m_{ij}} \subset (B_2)_{\mathfrak m_{ij}}$. Pick a maximal ideal $\mathfrak m'_{ij} \subset B_2$ lying over $\mathfrak m_{ij} \subset B_1$ (of course there may be more than one) and consider $$ \xymatrix{ (B_1)_{\mathfrak m_{ij}} \ar[r] & (B_2)_{\mathfrak m'_{ij}} \\ (A_1)_{\mathfrak m_i} \ar[u] \ar[r] & (A_2)_{\mathfrak m'_i} \ar[u] } $$ where $\mathfrak m'_i \subset A_2$ is the image. Now the horizontal arrows have ramification index $1$ and induce finite separable residue field extensions. Thus, using the equivalence of Lemma \ref{lemma-extension-dvrs-formally-smooth}, we see that it suffices to show that the right vertical arrow is formally smooth in the $\mathfrak m'_{ij}$-adic topology. Since $u$ has a $e$th root in $K_2$ we obtain the claim. \medskip\noindent Assume $u$ has an $e$th root in $K_1$. Since $e | e_1$ and since $u$ has a $e$th root in $K_1$ we see that $\pi = \theta^e$ for some $\theta \in K_1$. Let $K[\theta] \subset K_1$ be the subfield generated by $\theta$. By Lemma \ref{lemma-pull-root-uniformizer} the integral closure of $A$ in $K[\theta]$ is the discrete valuation ring $A[\theta]$. If we can prove the lemma for the extension $K \subset K[\theta]$, then $K \subset K[\theta]$ is a solution for $A \subset B$ and we conclude by Lemma \ref{lemma-formally-smooth-goes-up}. \medskip\noindent Assume $K_1 = K[\pi^{1/e}]$ and set $\theta = \pi^{1/e}$. Let $\pi_B$ be a uniformizer for $B$ and write $\pi = w \pi_B^e$ for some unit $w$ of $B$. Then we see that $L_1 = L \otimes_K K_1$ is obtained by adjoining $\pi_B/\theta$ which is an $e$th root of the unit $w$. Thus $B \subset B_1$ is finite \'etale. Thus for any maximal ideal $\mathfrak m \subset B_1$ consider the commutative diagram $$ \xymatrix{ B \ar[r]_1 & (B_1)_{\mathfrak m} \\ A \ar[u]^e \ar[r]^e & A_1 \ar[u]_{e_\mathfrak m} } $$ Here the numbers along the arrows are the ramification indices. By multiplicativity of ramification indices (Lemma \ref{lemma-multiplicative-e-f}) we conclude $e_\mathfrak m = 1$. Looking at the residue field extensions we find that $\kappa(\mathfrak m)$ is a finite separable extension of $\kappa_B$ which is separable over $\kappa_A$. Therefore $\kappa(\mathfrak m)$ is separable over $\kappa_A$ which is equal to the residue field of $A_1$ and we win by Lemma \ref{lemma-extension-dvrs-formally-smooth}.