This is from JS Milne's notes on Galois theory

Let $E/F$ be a Galois field extension (not necessarily finite). For any finite set $S \subseteq E$, let $G(S)$ be the subgroup of $Gal(E/F)$ that fixes $S$. 

Now suppose that $S$ is stable under $Gal(E/F)$, that is: for any $\sigma \in Gal(E/F)$ and $s \in S$, $\sigma(s) \in S$. Then, $G(S)$ is the kernel of the natural homomorphism $Gal(E/F) \rightarrow Sym(S)$. Therefore,  $G(S)$ is a normal subgroup and has finite index.

Take $Gal(E/F) \rightarrow \prod_{S \text{ finite, stable under $Gal(E/F)$}} Gal(E/F) / G(S)$ which is injective. Give  $Gal(E/F) / G(S)$ the discrete topology, and by Tychonoff's theorem, the right hand side is compact.

 Let $S_1 \subseteq S_2$ be stable sets. There are two group homomorphisms $ \prod_{S \text{ finite, stable under $Gal(E/F)$}} Gal(E/F) / G(S) \rightarrow Gal(E/F) / G(S_1)$. We can take the projection, or we can project to $Gal(E/F) / G(S_2)$ and then quotient, since $G(S_2) \subseteq G(S_1)$.

The next part is where I don't understand. Define $E(S_1, S_2)$ to be the subset of $\prod_{S \text{ finite, stable under $Gal(E/F)$}} Gal(E/F) / G(S)$ where the two homomorphisms are the same. The notes say that this is a closed set, and the intersection is the image of the morphism from $Gal(E/F)$. How do we prove these two statements?