Let $G$ be a profinite group, $A,B$ discrete $G$-modules and $A$ finitely generated. Then $\operatorname{Hom}_{\mathbb{Z}}(A,B)$ is a discrete $G$-module.
I don't see where exactly we need that $A$ is finitely generated so I'd be happy if anyone could tell me where the mistake in my argument is:
$\operatorname{Hom}_{\mathbb{Z}}(A,B)$ being a discrete $G$-module is equivalent to $\operatorname{Hom}_{\mathbb{Z}}(A,B) = \bigcup_{U \leq G\; \text{open}} \operatorname{Hom}_{\mathbb{Z}}(A,B)^U$. Since $A$ and $B$ are discrete, the corresponding equality holds for $A$ and $B$. Let $f: A \to B$ be a $\mathbb{Z}$-homomorphism and $u \in U $ where $U \leq G$ is some open subgroup. By the definition of the $G$-action on $\operatorname{Hom}_{\mathbb{Z}}(A,B)$ we have for all $a \in A$ $$(uf)(a) = uf(u^{-1}a) = f(u^{-1}a) = f(a),$$ hence $f \in \operatorname{Hom}_{\mathbb{Z}}(A,B)^U$ which is what we wanted to show.
The problem lies in the fact that even though $A = \bigcup_{U} A^U$, it is not true that $A = A^U$ for any particular open $U$. Thus, you need to be more careful about choosing your $U$ for $f \in Hom_{\mathbb{Z}}(A,B)$.
To use f.g., take generators $a_i$ of $A$. Then for each $i$, there exists open $U_i\subset G$ that fix $a_i$. Now take $U_A = \bigcap_i U_i$. By F.G, the intersection is finite, so $U_A$ is open. Now repeat this for $f(a_i)$, getting an open $U_B$ that fix all $f(a_i)$. Finally apply your argument to $U_A \cap U_B$.
Edit: you need $U_A$ to be open and normal, but this is guaranteed by the fact that $G$ is profinite.