Suppose $G$ is a compact (Polish, if it helps) group and $H\leq G$ is of countable index. Does it follow that the index is finite? If the answer is yes, what if we only know that the index is just strictly less than $2^{\aleph_0}$?
Equivalently, can a compact Hausdorff group act transitively on a countably infinite set? (Or an infinite set of cardinality below $2^{\aleph_0}$.)
I can show it under additional assumption that $H$ has the property of Baire (in this case, index under $2^{\aleph_0}$ implies that $H$ is clopen). On the other hand, it is not that hard to find examples where $H$ does not have the property of Baire (e.g. the kernel of a discontinuous functional on an infinite-dimensional vector space over a finite field), but the examples I know are all of finite index.
It doesn't follow that $H$ has to have finite index.
For example you can look at This paper Theorem 2.3
"Every infinite compact group G that is not profinite has a subgroup of countable index and thus contains a non-measurable subgroup."