Given $(A_i)_{i\in I}$ and $B$ modules over a ring $R$, with $I$ infinite, $A_i\ne 0\;\forall i$ and $B\ne 0$, is it true that
$$Hom_R(A_i,B)=0\;\forall i\implies Hom_R(\Pi_{i\in I} A_i, B)=0\quad?$$
I couldn't find a simple counterexample (Someone told me I can do it with $B=\mathbb{C}$ and $A_i$ finite fields, but I'm looking for something simpler)
This is a bit simpler, but along the exact same lines as the example you mention in your question.
For any module $M$ let $$tM = \{m \in M \ | \ m \ \text{has finite additive order}\}$$ be the torsion part of $M$. It's easy to see that $M/tM$ is torsion free, i.e., $t(M/tM) = 0$ and that $\mathbb Z/n$ is torsion, i.e., $t(\mathbb Z/n) = \mathbb Z/n$. It's also easy to see that there are no maps from a torsion module to a torsion free module.
Now let $$M = (\mathbb Z/2) \times (\mathbb Z/3) \times (\mathbb Z/4) \times \cdots$$ Note that $M/tM$ is nonzero because the element $(1, 1, 1, \ldots) \in M$ does not have finite additive order. Thus $B = M/tM$ is a nonzero module with the property that there are no maps $\mathbb Z/n \to B$ but there certainly is a map $M \to B$.