Let $L/K$ be a Galois extension and $E/K$ arbitrary (both contained in some extension of $K$), I want to prove that $\operatorname{Gal}(LE/E)\cong\operatorname{Gal}(L/E\cap L)$ as topological groups (wrt the Krull topology). Here is my attempt.
The extension $L/K$ is Galois $\implies L$ splitting field over $K$ of family of separable polynomials $\implies LE$ splitting field of said family over $E\implies LE/E$ Galois. Define $\rho:\operatorname{Gal}(LE/E)\to\operatorname{Gal}(L/K)$ via $\sigma\mapsto\sigma|_L$ ($L/K$ is normal, so $\sigma$ restricts to a $K$-automorphism of $L$), I want to show this is an open monomorphism of topological groups. This would imply that it's a homeomorphism onto its image, which will be shown to be $\operatorname{Gal}(L/E\cap L)$.
The basis of the topology on $\operatorname{Gal}(L/K)$ are cosets of $\operatorname{Gal}(L/F)$ for $F/K$ finite, it is known that this topology is compact and Hausdorff. The map $\rho$ is clearly an injective homomorphism of groups, we show it's continuous. It is enough to show continuity at the identity, i.e. that $\rho^{-1}(\operatorname{Gal}(L/K'))$ for $K'/K$ finite contains $\operatorname{Gal}(LE/E')$ for some $E'/E$ finite. Take $E'=EK'$, then $E'/E$ is finite and elements of $\operatorname{Gal}(LE/E)$ that leave $E'$ fixed are mapped by $\rho$ into $\operatorname{Gal}(L/K')\implies\rho$ is continuous. I don't know how to show $\rho$ is an open map.
We show the image $H$ of $\operatorname{Gal}(LE/E)$ is $\operatorname{Gal}(L/E\cap L)$. Since $\rho$ is continuous, $\operatorname{Gal}(LE/E)$ is compact, and $\operatorname{Gal}(L/K)$ is Hausdorff, we have that $H$ is a closed subgroup of $\operatorname{Gal}(LE/E)$; let $F=\operatorname{Fix}H$ be its fixed field. Clearly $F\supset E\cap L$. Conversely $F\subset\operatorname{Fix}\operatorname{Gal}(LE/E)=E\implies F=E\cap L$. But $E\cap L$ is the fixed field of $\operatorname{Gal}(L/E\cap L)$ (the extension $L/E\cap L$ is Galois), so $\operatorname{Fix}H=\operatorname{Fix}\operatorname{Gal}(L/E\cap L)$. Both of these are closed subgroups of $\operatorname{Gal}(L/K)$, therefore by the correspondence for infinite Galois extensions (intermediate fields correspond with closed subgroups) $H=\operatorname{Gal}(L/E\cap L)$.
The only part I'm having trouble with is showing $\rho$ is open. How do I do this? Is the rest of the proof correct?
To show that $\rho$ has continuous inverse, recall that every Galois group is compact Hausdorff. Consequently, $\rho$ is closed because: