One way to show this could be to first show that if $\alpha_1,\dots,\alpha_n \in L$ are algebraic and separable over $K$, then $K(\alpha_1,\dots,\alpha_n)$ is a separable extension of $K$ (that is, every element of $K(\alpha_1,\dots,\alpha_n)$ is separable over $K$). From this, it would follow that $$K_s = \bigcup K(\alpha_1,\dots,\alpha_n),$$ where the union is taken over all finite subsets $\{\alpha_1,\dots,\alpha_n\}$ of $L$ that consist of algebraic and separable elements over $F$. Then, one can show that $K_s$ is a field by a similar argument as used in showing that algebraic elements are closed under addition and multiplication.
One can use Galois theory to show that if $\alpha_1,\dots,\alpha_n \in L$ are algebraic and separable over $K$, then $K(\alpha_1,\dots,\alpha_n)$ is a separable extension of $K$. You can take a look at this answer of mine for a proof that uses Galois theory.