Let $G$ be a profinite subgroup. And $K ,H$ be closed subgroups of $G$ such that
$K \vartriangleleft H$ and $(H:K) < \infty $ .
$K \subseteq B \subseteq H$ (subgroups)
Then $B$ is closed in $G$?
This question originates from the following statement in Neukirch, Algebraic Number Theory, p.301 (We assume knowledge of Abstract Galois Theory in his book. If additional explanation is needed, then I am willing to.)
My tiral for understanding is the underlined statement. I understand the underlined statement as finding a subextension $M|K$(meaning as Neukirch's book p.275) of $L|K$ such that $G(L|M)$ ($L|M$ is Galois) is a p-Sylow subgroup of $G(L|K)$. Then my argument is as follows ;
Attempt : We are given a profinite group $G$ (c.f. Neukirch's book, p.275) and finite galois extension $L|K$ ; that is, indices of closed subgroups $G_L$, $G_K$ of $G$ such that $G_L$ is normal in $G_K$ and $(G_K : G_L)$ is finite. Now, let $S_p \subseteq G(L|K):=G_K/G_L$ a $p$-Sylow subgroup and $\pi : G_K \to G(L|K):=G_K/G_L$ be the natural proejction. Let $B:= \pi ^{-1}(S_p)$. Then, note that $$G_L \subseteq B \subseteq G_K$$
So, if $B$ is closed in $G$, then we let $G_M := B$ (c.f. also p.275 in his book). Then $M|K$ is a subextension of $L|K$ and we are done. So is $B$ really closed in $G$?
$[B:K] < \infty$, so $B$ is the union of a finite number of translations of $K$, which is closed, so $B$ is closed.