Let $k$ be a finite field. I am trying to prove that the absolute Galois group of $k$, i.e., $G = \operatorname{Gal}(\bar{k} / k)$ where $\bar{k}$ is an algebraic closure of $k$, is strongly complete, i.e., each subgroup $H$ of $G$ of finite index is open in the Krull topology of $G$.
Since $G$ is a disjoint union of finitely many left (or right) cosets of $H$, and since the left (or right) multiplication for a fixed element of $G$ is an homeomorphism, I found that $H$ is open if and only if it is closed. Moreover, from the Galois correspondence I know that $\operatorname{Gal}(\bar{k}/\bar{k}^H) = \operatorname{cl}(H)$, where $\bar{k}^H$ is the subfield of $\bar{k}$ fixed by $H$ and $\operatorname{cl}(H)$ is the topological closure of $H$.
However, I can find no way to prove that $H = \operatorname{cl}(H)$. Probably I am missing something trivial.
Thank you in advance for any suggestion.
Before starting, let me note that what you have asked is just a particular case of the Nikolov-Segal's theorem (On finitely generated profinite groups I and II) that states that in a finitely generated profinite group, every subgroup of finite index is open. This was a generalization of a previous result by Serre for finitely generated pro-$p$ groups.
So this means that the important think is that the profinite group $G$ that we are considering is finitely generated. In particular, it is procyclic so thinks get even easier.
The first we should note is that $$G^n=\{g^n\,|\,g\in G\}$$ is a closed subgroup of $G$ since it is the image of $G$ under the continuous homomorphism of groups $g\mapsto g^n$. Furthermore, in our considered case --the procylic group generated by the Fröbenius automorphism--, we can easily see that $$|G:G^n|=n$$ Now, given a subgroup $H$ of $G$ of finite index $n$, we should have $$G^n\leq H$$ since the $n$-th power of every element of $G$ is the identity in the quotient $G/H$. But $G^n$ and $H$ have the same index, so they must be equal and therefore $H$ is closed as desired.