The separable degree is independent of the chosen algebraic closure

Choose a finite $F$-basis ${\alpha }_1,\dots ,{\alpha }_r$ of $K$ by [L2], and let $f$ be the product of their minimal polynomials over $F$ from [L3]. For $j=1,2$, let $E_j\subseteq {\mathrm{\Omega }}_j$ be generated over $F$ by all roots of $f$ in ${\mathrm{\Omega }}_j$. Since ${\mathrm{\Omega }}_j$ is algebraically closed, $f$ splits there, and [L4] makes $E_j\mathrm{/}F$ a splitting field of $f$.

By [L5], choose an $F$-isomorphism $\theta :E_1\to E_2$. Postcomposition with $\theta$ gives a bijection ${\mathrm{Hom}}_F(K,E_1)\to {\mathrm{Hom}}_F(K,E_2)$, with inverse given by postcomposition with ${\theta }^{-1}$.

Every $F$-embedding $\sigma :K\to {\mathrm{\Omega }}_j$ sends each ${\alpha }_i$ to a root of its minimal polynomial, so $\sigma (K)=F(\sigma ({\alpha }_1),\dots ,\sigma ({\alpha }_r))\subseteq E_j$. Hence ${\mathrm{Hom}}_F(K,{\mathrm{\Omega }}_j)={\mathrm{Hom}}_F(K,E_j)$.

Steps 2.1 and 1.2 give a bijection between the embedding sets into ${\mathrm{\Omega }}_1$ and ${\mathrm{\Omega }}_2$. Their finite cardinalities are equal, so the value in [L1] is independent of the closure.