Let $M'$ be any finite extension of $L$ and consider the integral closure $C'$ of $B$ in $M'$. Then $C'$ is finite over $B$ as $B$ is Nagata by Algebra, Lemma \ref{algebra-lemma-Noetherian-complete-local-Nagata}. Moreover, $C'$ is a discrete valuation ring, see discussion in Remark \ref{remark-construction}. Moreover $C'$ is complete as a $B$-module, hence complete as a discrete valuation ring, see Algebra, Section \ref{algebra-section-completion}. It follows in particular that $C$ is the integral closure of $B$ in $M$ (by definition of valuation rings as maximal for the relation of domination). \medskip\noindent Let $M \subset M'$ be a finite extension and let $C' \subset M'$ be the integral closure of $B$ as above. By Lemma \ref{lemma-solutions-go-down} it suffices to prove the result for $A \to B \to C'$. Hence we may assume that $M/L$ is normal, see Fields, Lemma \ref{fields-lemma-normal-closure}. \medskip\noindent If $M / L$ is normal, we can find a chain of finite extensions $$ L = L^0 \subset L^1 \subset L^2 \subset \ldots \subset L^r = M $$ such that each extension $L^{j + 1}/L^j$ is either: \begin{enumerate} \item[(a)] purely inseparable of degree $p$, \item[(b)] totally ramified with respect to $B^j$ and Galois of degree $p$, \item[(c)] totally ramified with respect to $B^j$ and Galois cyclic of order prime to $p$, \item[(d)] Galois and unramified with respect to $B^j$. \end{enumerate} Here $B^j$ is the integral closure of $B$ in $L^j$. Namely, since $M/L$ is normal we can write it as a compositum of a Galois extension and a purely inseparable extension (Fields, Lemma \ref{fields-lemma-normal-case}). For the purely inseparable extension the existence of the filtration is clear. In the Galois case, note that $G$ is ``the'' decomposition group and let $I \subset G$ be the inertia group. Then on the one hand $I$ is solvable by Lemma \ref{lemma-galois-inertia} and on the other hand the extension $M^I/L$ is unramified with respect to $B$ by Lemma \ref{lemma-inertial-invariants-unramified}. This proves we have a filtration as stated. \medskip\noindent We are going to argue by induction on the integer $r$. Suppose that we can find a finite extension $K \subset K_1$ which is a weak solution for $A \to B^1$ where $B^1$ is the integral closure of $B$ in $L^1$. Let $K'_1$ be the normal closure of $K_1/K$ (Fields, Lemma \ref{fields-lemma-normal-closure}). Since $A$ is complete and the residue field of $A$ is algebraically closed we see that $K'_1/K_1$ is separable and totally ramified with respect to $A_1$ (some details omitted). Hence $K \subset K'_1$ is a weak solution for $A \to B^1$ as well by Lemma \ref{lemma-weakly-unramified-goes-up-along-totally-ramified}. In other words, we may and do assume that $K_1$ is a normal extension of $K$. Having done so we consider the sequence $$ L^0_1 = (L^0 \otimes_K K_1)_{red} \subset L^1_1 = (L^1 \otimes_K K_1)_{red} \subset \ldots \subset L^r_1 = (L^r \otimes_K K_1)_{red} $$ and the corresponding integral closures $B^i_1$. Note that $C_1 = B^r_1$ is a product of discrete valuation rings which are transitively permuted by $G = \text{Aut}(K_1/K)$ by Lemma \ref{lemma-galois-relative}. In particular all the extensions of discrete valuation rings $A_1 \to (C_1)_\mathfrak m$ are isomorphic and a solution for one will be a solution for all of them. We can apply the induction hypothesis to the sequence $$ A_1 \to (B^1_1)_{B^1_1 \cap \mathfrak m} \to (B^2_1)_{B^2_1 \cap \mathfrak m} \to \ldots \to (B^r_1)_{B^r_1 \cap \mathfrak m} = (C_1)_\mathfrak m $$ to get a solution $K_1 \subset K_2$ for $A_1 \to (C_1)_\mathfrak m$. The extension $K \subset K_2$ will then be a solution for $A \to C$ by what we said before. Note that the induction hypothesis applies: the ring map $A_1 \to (B^1_1)_{B^1_1 \cap \mathfrak m}$ is weakly unramified by our choice of $K_1$ and the sequence of fraction field extensions each still have one of the properties (a), (b), (c), or (d) listed above. Moreover, observe that for any finite extension  $\kappa_B \subset \kappa$ we still have $k = \bigcap \kappa^{p^n}$. \medskip\noindent Thus everything boils down to finding a weak solution for $A \subset C$ when the field extension $L \subset M$ satisfies one of the properties (a), (b), (c), or (d). \medskip\noindent Case (d). This case is trivial as here $B \to C$ is unramified already. \medskip\noindent Case (c). Say $M/L$ is cyclic of order $n$ prime to $p$. Because $M/L$ is totally ramified with respect to $B$ we see that the ramification index of $B \subset C$ is $n$ and hence the ramification index of $A \subset C$ is $n$ as well. Choose a uniformizer $\pi \in A$ and set $K_1 = K[\pi^{1/n}]$. Then $K_1/K$ is a solution for $A \subset C$ by Abhyankar's lemma (Lemma \ref{lemma-abhyankar}). \medskip\noindent Case (b). We divide this case into the mixed characteristic case and the equicharacteristic case. In the equicharacteristic case this is Lemma \ref{lemma-characteristic-p-case}. In the mixed characteristic case, we first replace $K$ by a finite extension to get to the situation where $M/L$ is a degree $p$ extension of finite level using Lemma \ref{lemma-make-finite-level}. Then the level is a rational number $l \in [0, p)$, see discussion preceding Lemma \ref{lemma-lowering-the-level}. If the level is $0$, then $B \to C$ is weakly unramified and we're done. If not, then we can replacing the field $K$ by a finite extension to obtain a new situation with level $l' \leq \max(0, l - 1, 2l - p)$ by Lemma \ref{lemma-lowering-the-level}. If $l = p - \epsilon$ for $\epsilon < 1$ then we see that $l' \leq p - 2\epsilon$. Hence after a finite number of replacements we obtain a case with level $\leq p - 1$. Then after at most $p - 1$ more such replacements we reach the situation where the level is zero. \medskip\noindent Case (a) is Lemma \ref{lemma-purely-inseparable-case}. This is the only case where we possibly need a purely inseparable extension of $K$, namely, in case (2) of the statement of the lemma we win by adjoining a $p$th power of the element $\pi$. This finishes the proof of the lemma.