Let $G$ be a profinite group. It is known that open normal subgroups of $G$ have finite index, but in general not every normal subgroup of finite index is open (see https://ncatlab.org/nlab/show/profinite+group#examples for a counterexample). This last fact is true if the profinite group is (topologically) finitely generated (this is a theorem of Nikolov–Segal).
Let now $K$ be a field. A theorem about the "inverse Galois problem" states that for every profinite group $G$ there are fields $K\subseteq L\subseteq F$ such that $Gal(F/L)\cong G$. This means that also for general Galois groups, it is not true that every normal subgroup of finite index is open.
My question is: could this be always true for absolute Galois groups, i.e. for Galois groups of extensions $\bar{K}/K$ with $K$ field and $\bar{K}$ the separable closure of $K$?
My question is justified by a sentence I found in Silverman's book "The Arithmetic of Elliptic Curves" where, at the beginning of the appendix B.2, it is written
"Let $G_{\bar{K}/K}$ be the Galois group of $\bar{K}/K$. (...) Thus $G_{\bar{K}/K}$ is a profinite group, i.e. an inverse limit of finite groups. As such, it comes equipped with a topology in which a basis of open sets around the identity consists of the collection of normal subgroups having finite index in $G_{\bar{K}/K}$".
The paragraph you cited from Silverman is referring to the following fact: In a profinite group, the collection of open normal subgroups form a base of neighborhoods of the identity. (Lemma 1-17, p.29 in Fourier Analysis on Number Fields).
Here on a Galois group, Silverman considers the Krull topology. Concerning the stabilizer, I think he should require the stabilizer to be open rather than just of finite index.