Let $G$ be a profinite group.
Question: Is there always a chain of open normal subgroups $G=G_1\geq G_2\geq \cdots$ of $G$ such that $\cap G_i=\{1\}$? If not, is there some useful sufficient condition to ensure this?
I know that the identity element $1$ of $G$ admits a fundamental system $\{U_i\}$ of open neighborhoods such that $\cap U_i=\{1\}$ and each $U_i$ is open normal subgroup of $G$. The motivation is from the following question about Galois theory: If we have an infinite Galois group $\text{Gal}(L/K)$ for some field extension $L/K$, then we can find a chain of field extension $K\subset K_1\subset \cdots \subset K_i\subset \cdots\subset L$ such that $K_i/K$ is finite Galois extension and $|\text{Gal}(K_i/K)|\to \infty$ as $i\to \infty$ assuming that the answer to the above question is Yes.
No, such a chain does not always exist. For a simple counterexample, let $F$ be any nontrivial finite group and $G=F^I$ for some uncountable set $I$. Then any open subset of $G$ is unconstrained on all but finitely many coordinates and so $\{1\}$ cannot be written as an intersection of countably many open sets.
If you know that $G$ is first-countable then it is true, since then there is a neighborhood base at $1$ consisting of countably many open normal subgroups and by taking intersections of these you can get a nested sequence.
As for your application, you need something much weaker than $\bigcap G_i=\{1\}$; you just need $|G/G_i|$ to be unbounded. This will be true as long as $G_{i+1}$ is strictly contained in $G_i$ for each $i$, so that the index in $G$ strictly increases each time. This can easily be achieved (assuming $G$ is infinite), since you can just take any $g\in G_i\setminus\{1\}$, let $U$ be an open normal subgroup that does not contain $g$, and let $G_{i+1}=G_i\cap U$.