Proof.
The lemma is trivial if the characteristic of $K$ is zero; thus we may and do assume that the characteristic of $K$ is $p > 0$.
Let $K_2/K$ be a solution for $A \to B$. We will use induction on the inseparable degree $[K_2 : K]_ i$ (Fields, Definition 9.14.7) of $K_2/K$. If $[K_2 : K]_ i = 1$, then $K_2$ is separable over $K$ and we are done. If not, then there exists a subfield $K_2/K_1/K$ such that $K_2/K_1$ is purely inseparable of degree $p$ (Fields, Lemmas 9.14.6 and 9.14.5). By Lemma 15.116.5 there exists a separable extension $K_3/K_1$ which is a solution for $A \subset B$. Then $[K_3 : K]_ i = [K_1 : K]_ i = [K_2 : K]_ i/p$ (Fields, Lemma 9.14.9) is smaller and we conclude by induction.
$\square$
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