Proof. sketch (TODO) [17011A]

$\implies$: Suppose $L / F$ is separable. Given $\alpha \in K$, we have that $\alpha \in L$ is hence separable over $F$, so $K / F$ is separable. To show that $L / K$ is separable, consider $m_{\alpha, F}(x)$. Since $\alpha$ is separable over $F$, all roots of $m_{\alpha, F}(x)$ are simple. Moreover, $m_{\alpha, K}(x) \mid m(\alpha, F)(x)$ in $K[x]$, so all roots of $m_{\alpha, K}(x)$ are simple as well $\implies$ $\alpha$ is separable over $K$. Hence $L$ is separable over $K$.

$\impliedby$: Suppose that $K / F$ and $L / K$ are both separable.

Case 1: $[L : F] < \infty$. By Proposition 1707, $$ \#\{ F\text{-embeddings }\varphi : K \to \overline{F} \} = [K : F], \, \#\{ K\text{-embeddings } \varphi : L \to \overline{K} \} = [L : K]. $$ Then Proposition 17010 implies $$ \#\{ F\text{-embeddings }\varphi : L \to \overline{F} \} = [K : F][L : K] = [L : F]. $$ By Proposition 1707, this implies that $L / F$ is separable.

Case 2: $[L : F] = \infty$.