Here is the statement of the problem:
Let $K/F$ be Galois with $G = \mathrm{Gal}(K/F)$, and let $L$ be an intermediate field. Let $N \subseteq K$ be the normal closure of $L/F$. If $H = \mathrm{Gal}(K/L)$, show that $\mathrm{Gal}(K/N) = \bigcap_{\sigma \in G} \sigma H \sigma^{-1}$.
I am able to show that $\mathrm{Gal}(K/N) \subseteq \bigcap_{\sigma \in G} \sigma H \sigma^{-1}$ without too much trouble.
If I assume that $K/F$ is a finite extension, I can use the Fundamental Theorem of Galois Theory to show that $\mathrm{Gal}(K/N) = \bigcap_{\sigma \in G} \sigma H \sigma^{-1}$ as follows:
Note that $\bigcap_{\sigma \in G} \sigma H \sigma^{-1}$ is normal in $G$. We have that $\mathrm{Gal}(K/N) \subseteq \bigcap_{\sigma \in G} \sigma H \sigma^{-1} \subseteq \mathrm{Gal}(K/L)$. Then, let $\bigcap_{\sigma \in G} \sigma H \sigma^{-1}$ correspond to the intermediate field $M$ under the Galois correspondence, we have that $N \supseteq M \supseteq L$. Since $\bigcap_{\sigma \in G} \sigma H \sigma^{-1}$ is a normal subgroup of $G$, $M/F$ is Galois and therefore normal. Hence, $M/L$ is normal as well. Hence, either $M = N$ or $M = L$ by Proposition 5.9. If $M = N$, then $\bigcap_{\sigma \in G} \sigma H \sigma^{-1}= \mathrm{Gal}(K/N)$ by the FTGT, and we are done. If $M = L$, then $L/F$ is normal, implying that $L = N = M$, which brings us back to the previous case.
Is there another proof, or a modification of this one, that does not use the fact that $K/F$ is a finite extension? There is an answer to this question here, but it uses the following reasoning: since $N$ is the normal closure of $L/F$, there exists some $σ∈G$ such that $σ^{−1}(a)∈L$. Is this true? I don't see why, but if it is, then that would answer my question.