Assuming Choice, separable closures exist and are base-isomorphic

Choose an algebraic closure $\mathrm{\Omega }\mathrm{/}F$ by [L1] and let $S={\mathrm{\Omega }}_s$ be its relative separable closure from [L2]. Then $S\mathrm{/}F$ is algebraic and separable.

Let $S_1,S_2$ be separable closures and choose an algebraic closure ${\mathrm{\Omega }}_2$ of $S_2$ using [L1]. By [L5], the identity of $F$ extends to an embedding $\sigma :S_1\to {\mathrm{\Omega }}_2$. Images of elements separable over $F$ remain separable. Conversely, if $a\in {\mathrm{\Omega }}_2$ is separable over $F$, then its minimal polynomial over $S_2$ divides its separable minimal polynomial over $F$, so it is separable over the separably closed field $S_2$ and therefore lies in $S_2$. Hence $\sigma (S_1)\subseteq S_2$.

Let $q\in S[x]$ be nonconstant and separable. Its finitely many coefficients generate a finite subextension $E\mathrm{/}F$ of $S\mathrm{/}F$ by [L6]. By [L4], $q$ is already separable over $E$. Each root $a\in \mathrm{\Omega }$ is separable over $E$, while $E\mathrm{/}F$ is separable; [L3] makes $a$ separable over $F$, so $a\in S$. Thus $q$ splits in $S$.

The image $\sigma (S_1)$ is separably closed. For $a\in S_2$, its minimal polynomial over $\sigma (S_1)$ divides its separable minimal polynomial over $F$, so it is separable over $\sigma (S_1)$. Separable closedness therefore forces $a$ into the image. Thus $\sigma :S_1\to S_2$ is an $F$-isomorphism. The construction does not assert uniqueness.

Steps 1.1 and 2.1 give existence, while steps 1.2 and 2.2 give base-field isomorphism of any two separable closures.