**Update:** Further work with Adam (who answers below) and Piotr led to a rather satisfactory result about the problem that motivated the problem below, see our recent paper
The Haar Measure Problem. In particular, we answer there a problem mentioned in the discussion below.

The following question is motivated by the paper
[Brian, Mislove, *Every compact group can have a non-measurable subgroup*].
A positive solution to a variation of the following problem implies a positive
solution to the problem studied there, i.e., that every infinite compact group has a subgroup that is not Haar measurable. A more general form of this question was answered in the negative there.

Every infinite metric profinite group is, as a topological space, homeomorphic to
the Cantor space [Hofmann, Morris, **The Structure of Compact Groups**, Theorem 10.40].

**Problem.**
Let $G$ be an infinite metric profinite group (or, more generally, one homeomorphic to the Cantor space), $H$ a countable subgroup of $G$, and $g\in G\smallsetminus H$.
Is there necessarily an element $x\in G\smallsetminus H$ such that $g\notin\langle H\cup \{x\}\rangle$?

**Discussion.** I find it intriguing that in this scenario, the problem is about a topological group structure on the Cantor space. A rather well-understood object from the topological point of view. In particular, this fact is used in the Brian--Mislove paper cited above to observe that the answer to the big question in the first paragraph is positive when there is a non-null set of reals of cardinality smaller than the continuum ($\mathrm{non}(\mathcal{N})<\mathfrak{c}$). Thus, it remains to consider the case $\mathrm{non}(\mathcal{N})=\mathfrak{c}$ and a transfinite induction can be used to construct the needed nonmeasurable subgroup. There, in each step, the intermediate subgroup is Lebesgue null. In our problem, we model this by "countable".
Even assuming the Continuum Hypothesis, I do not know the answer to Problem 2 (or the big problem).

Comment: I have separated this question from the quetion there to make it possible to declare the other question as solved.