In the A taste of Topology book, when talking about Cartesian product $\prod\{S:S\in\mathcal{S}\}$, the author writes the following:
It is straightforward that $\prod\{S:S\in\mathcal{S}\}\neq\emptyset$ whenever $\mathcal{S}$ is finite and $S\neq\emptyset$ for any $S\in\mathcal{S}$. The same can be shown if $\mathcal{S}$ is countable, see Exercise 2.
The "Exercise 2" asks to prove that $\prod_n S_n$ of non-empty sets is non-empty, without invoking Zorn's lemma.
Is it indeed that straightforward to show the claim for countable $\mathcal{S}$? I was under the impression that while the full strength of AC is not needed, one still needs the weaker Axiom of Countable Choice to make that claim. Can it be proved in ZF alone?
Based on the information you provide this is a mistake. You are correct in that it is consistent that there is a countable family of finite sets whose product is empty (namely, they do not have a choice function).
For example in Cohen's second model for the failure of choice there is such family of sets of size $2$.