Observe that a weak solution is a solution if the residue field of $A$ is perfect, see Lemma \ref{lemma-extension-dvrs-formally-smooth}. Thus the proposition follows immediately from Theorem \ref{theorem-epp} if the residue characteristic of $A$ is $0$ (and in fact we do not need the assumption that $A \to B$ is essentially of finite type). If the residue characteristic of $A$ is $p > 0$ we will also deduce it from Epp's theorem. \medskip\noindent Let $x_i \in A$, $i \in I$ be a set of elements mapping to a $p$-base of the residue field $\kappa$ of $A$. Set $$ A' = \bigcup\nolimits_{n \geq 1} A[t_{i, n}]/(t_{i, n}^{p^n} - x_i) $$ where the transition maps send $t_{i, n + 1}$ to $t_{i, n}^p$. Observe that $A'$ is a filtered colimit of weakly unramified finite extensions of discrete valuation rings over $A$. Thus $A'$ is a discrete valuation ring and $A \to A'$ is weakly unramified. By construction the residue field $\kappa' = A'/\mathfrak m_A A'$ is the perfection of $\kappa$. \medskip\noindent Let $K'$ be the fraction field of $A'$. We may apply Lemma \ref{lemma-big-extension-is-ok} to the extension $K \subset K'$. Thus $B'$ is a finite product of Dedekind domains. Let $\mathfrak m_1, \ldots, \mathfrak m_n$ be the maximal ideals of $B'$. Using Epp's theorem (Theorem \ref{theorem-epp}) we find a weak solution $K' \subset K'_i$ for each of the extensions $A' \subset B'_{\mathfrak m_i}$. Since the residue field of $A'$ is perfect, these are actually solutions. Let $K' \subset K'_1$ be a finite extension which contains each $K'_i$. Then $K' \subset K'_1$ is still a solution for each $A' \subset B'_{\mathfrak m_i}$ by Lemma \ref{lemma-formally-smooth-goes-up}. \medskip\noindent Let $A'_1$ be the integral closure of $A$ in $K'_1$. Note that $A'_1$ is a Dedekind domain by the discussion in Remark \ref{remark-construction} applied to $K' \subset K'_1$. Thus Lemma \ref{lemma-big-extension-is-ok} applies to $K \subset K'_1$. Therefore the integral closure $B'_1$ of $B$ in $L'_1 = (L \otimes_K K'_1)_{red}$ is a Dedekind domain and because $K' \subset K'_1$ is a solution for each $A' \subset B'_{\mathfrak m_i}$ we see that $(A'_1)_{A'_1 \cap \mathfrak m} \to (B'_1)_{\mathfrak m}$ is formally smooth in the $\mathfrak m$-adic topology for each maximal ideal $\mathfrak m \subset B'_1$. \medskip\noindent By construction, the field $K'_1$ is a filtered colimit of finite extensions of $K$. Say $K'_1 = \colim_{i \in I} K_i$. For each $i$ let $A_i$, resp.\ $B_i$ be the integral closure of $A$, resp.\ $B$ in $K_i$, resp.\ $L_i = (L \otimes_K K_i)_{red}$. Then it is clear that $$ A'_1 =  \colim A_i\quad\text{and}\quad B'_1 = \colim B_i $$ Since the ring maps $A_i \to A'_1$ and $B_i \to B'_1$ are injective integral ring maps and since $A'_1$ and $B'_1$ have finite spectra, we see that for all $i$ large enough the ring maps $A_i \to A'_1$ and $B_i \to B'_1$ are bijective on spectra. Once this is true, for all $i$ large enough the maps $A_i \to A'_1$ and $B_i \to B'_1$ will be weakly unramified (once the uniformizer is in the image). It follows from multiplicativity of ramification indices that $A_i \to B_i$ induces weakly unramified maps on all localizations at maximal ideals of $B_i$ for such $i$. Increasing $i$ a bit more we see that $$ B_i \otimes_{A_i} A'_1 \longrightarrow B'_1 $$ induces surjective maps on residue fields (because the residue fields of $B'_1$ are finitely generated over those of $A'_1$ by Lemma \ref{lemma-big-extension-is-ok}). Picture of residue fields at maximal ideals lying under a chosen maximal ideal of $B'_1$: $$ \xymatrix{ \kappa_{B_i} \ar[r] & \kappa_{B_{i'}} \ar[r] &  & \ldots & \kappa_{B'_1} \\ \kappa_{A_i} \ar[r] \ar[u] & \kappa_{A_{i'}} \ar[r] \ar[u] &  & \ldots & \kappa_{A'_1} \ar[u] } $$ Thus $\kappa_{B_i}$ is a finitely generated extension of $\kappa_{A_i}$ such that the compositum of $\kappa_{B_i}$ and $\kappa_{A'_1}$ in $\kappa_{B'_1}$ is separable over $\kappa_{A'_1}$. Then that happens already at a finite stage: for example, say $\kappa_{B'_1}$ is finite separable over $\kappa_{A'_1}(x_1, \ldots, x_n)$, then just increase $i$ such that $x_1, \ldots, x_n$ are in $\kappa_{B_i}$ and such that all generators satisfy separable polynomial equations over $\kappa_{A_i}(x_1, \ldots, x_n)$. This means that $A_i \to (B_i)_\mathfrak m$ is formally smooth in the $\mathfrak m$-adic topology for all maximal ideals $\mathfrak m$ of $B_i$ and the proof is complete.