Proof. sketch [22020A]

    1. Given any $\alpha' \in S'$, by construction, $\alpha'$ is a root of $m_{\alpha, F}(x)$ for some $\alpha \in S$. And clearly $m_{\alpha', F}(x) = m_{\alpha, F}(x)$. Since $K = F(S) / F$ is separable, $m_{\alpha, F}(x)$ is separable over $F$ $\implies$ $m_{\alpha', F}(x)$ is separable over $F$ $\implies$ $F(S') / F$ is separable. Moreover, any other roots of $m_{\alpha', F}(x)$ lie in $S'$, hence in $F(S')$ $\implies$ $m_{\alpha', F}(x)$ splits completely over $F(S')$ $\implies$ $F(S') / F$ is normal. Therefore, $F(S') / F$ is Galois.

      Suppose $K \subseteq E \subseteq F(S')$ and $E / F$ is Galois. Then for all $\alpha \in S$, $m_{\alpha, F}(x)$ splits completely over $E$ $\implies$ all the roots of $m_{\alpha, F}(x)$ lie in $E$ $\implies$ $S' \subseteq E$ $\implies F(S') \subseteq E$ $\implies F(S') = E$.

      So $F(S')$ is a Galois closure of $K$ over $F$.

    2. Let $\varphi : L \to \overline{K} = \overline{F}$ be a $K$-embedding such that $\varphi |_{K} = \operatorname{id}_{K}$. So $K \subseteq \varphi(L) \subseteq \overline{F}$. Since $\varphi$ is injective, we have $\varphi(L) \cong L$ and we can check that $\varphi(L)$ is also a Galois closure of $K$ over $F$. Then by the minimality of Galois closure, $\varphi(L) = F(S')$. Hence $L \cong F(S')$.

    3. Since $K \subseteq F(S') \subseteq \overline{F}$ and $F(S') / F$ is Galois, we have $$F(S') \subseteq \bigcap_{\begin{aligned}K \subseteq L \subseteq \overline{F} \\ L / F \text{ Galois}\end{aligned}} L.$$ The other direction just follows from the construction of $S'$.