Fix $\overline{F} = \overline{K}$. Then $F \subseteq K \subseteq \overline{F}$.

$\implies$: Let $A = \{ m_{\alpha, F}(x) \in F[x] \mid \alpha \in K \}$. First, we show that $K$ is the splitting field of $A \subseteq F[x]$. Let $S = \{ \text{roots of } \overline{F} \text{ of polynomials in } A \}$. Then $F(S)$ is the splitting field for $A$. We need to show $K = F(S)$. Given $\alpha \in K$, by the definition of $A$, $m_{\alpha, F}(x) \in A$ and by the definition of $S$, $\alpha$ is a root of $m_{\alpha, F}(x) \in A$, so $\alpha \in S$. Thus, $K \subseteq S \subseteq F(S)$. Conversely, if $\beta \in S$, then $\beta$ is a root of $m_{\alpha, F}(x)$ for some $\alpha \in K$. Since $K / F$ is normal, $K$ contains all the roots of $m_{\alpha, F}(x)$ in $\overline{F}$, so $K$ contains $\beta$. This shows that $S \subseteq K$, so $F(S) \subseteq K$. We have proved $K = F(S)$, the splitting field for $A \subseteq F[x]$.

Also, because $K / F$ is separable, we know that every $\alpha \in K$ is separable over $F$ and $m_{\alpha, F}(x)$ is separable. Hence $A$ is a subset of separable polynomials in $F[x]$.

$\impliedby$: Suppose $A$ is the subset of separable polynomials in $F[x]$ and $K$ is the splitting field for $A \subseteq F[x]$. We show that $K / F$ is both normal and separable. Proposition 2205 $\implies$ $K / F$ is normal. By the proof of Proposition 2204, $K \cong F(S)$ where $S \subseteq \overline{F}$ is the set of roots of polynomials in $A$. The polynomials in $A$ are separable $\implies$ every element in $S$ is separable over $F$. Then Proposition 17012 $\implies$ $K = F(S)$ is a separable extension of $F$.