Let $A$ be a (discrete) countable group such that for ever completely metrizable group $M$ and any (not necessarily continuous) homomorphism $f\colon M \to A$ there exists some open normal subgroup $N\trianglelefteq M$ with finite image $f(N)$.
Now, let $G$ be an extension of $A$ by a finite (discrete) group. Does $G$ admit the same property above?
By the universal embedding theorem of Krasner–Kaloujnine it is enough to show that this holds for semidirect products of $A\rtimes F$, where $F$ is a finite group. Further, we might assume that for a homomorphism $g\colon M \to A\rtimes F$ every identity neighbourhood in $M$ maps surjectively onto $A\rtimes F$ (by replacing $M$ with some open normal subgroups with minimal image with respect to $M\to A\rtimes F \to F$). If (after those reductions) $F$ is trivial, we are done. But in the other case, I am completely stuck.