Assume for each sequence $(X_n)_{n\in \mathbb{N}}$ of two-element sets, $\displaystyle\prod_{n\in\mathbb{N}}X_n \neq \emptyset$. Can we then find a Gale-Stewart game that isn't determined ?
EDIT: Noah Schweber has answered the first question, but I realized I wanted to ask something a bit different, namely changing the index set to an arbitrary set, instead of $\mathbb{N}$ (but the $X_i$ are still two-element sets). What about it then ?
No, you cannot.
$AC(2)$ follows from dependent choice ($DC$), which is consistent with $AD$ (see e.g. this paper by Steel). But by $AD$, every Gale-Stewart game is determined. So $AC(2)$ isn't enough choice to build an undetermined Gale-Stewart game.
Incidentally, invoking DC here is overkill, but my point is that in order to build an undetermined game you need a lot of choice. In particular, most of "ordinary" mathematics can be done in ZF+DC; see e.g this question (which also points out that there is some ordinary mathematics not doable from ZF+DC alone).
Here's the proof that $DC$ implies $AC(2)$. Take your given instance $(X_i)_{i\in\omega}$ of $AC(2)$, and consider the set $A$ of finite partial maps $p$ from $\omega$ to $\bigcup X_i$ such that
$dom(p)\in\omega+1$ (that is, is an initial segment of $\omega$), and
$p(i)\in X_i$ for $i\in dom(p)$ (that is, $p$ is a partial choice function for $(X_i)_{i\in\omega}$).
Now apply $DC$ to $A$ with the relation $\subsetneq$. We get a sequence $p_0\subsetneq p_1\subsetneq . . . $ of elements of $A$; now show that $\bigcup p_i$ is a choice function for $(X_i)_{i\in\omega}$.