Since closed sets are Borel, it follows that the Axiom of Choice is a theorem of ZF. I am aware this is wrong and very likely an entirely trivial confusion emerging from my lack of knowledge on these matters. However, I'd appreciate someone to shed some light on this, probably with a line pointing out what is the confusion.
Thanks!
Closed, or Borel, in what space?
$\mathsf{ZF}$ proves that the axiom of choice is equivalent to the following statement:
$(*)\quad$ For every set $X$, equip $X^\omega$ with the product topology coming from the discrete topology on $X$. Then every game on $X$ of length $\omega$, whose payoff set is closed as a subset of $X^\omega$, is determined.
But $(*)$ is an incredibly general principle! Every set $X$ has to be taken into account here. By contrast, when we talk about Borel determinacy we're (unless otherwise stated!) referring to specifically Borel subsets of $\omega^\omega$ (or equivalently for this context, $2^\omega$). While $\mathsf{ZF}$ proves "Every Borel subset of $\omega^\omega$ is determined," this is not enough to imply $(*)$ above.
Indeed, in general the phrase "$\Gamma$-determinacy" refers either to $\omega^\omega$ or $2^\omega$ specifically (and even this needs further clarification for $\Gamma$ sufficiently simple - closed games on $2$ are fairly boring but closed games on $\omega$ are extremely complicated). So the phrase "Over $\mathsf{ZF}$, choice is equivalent to closed determinacy" is misleadingly phrased; it should be something like "Over $\mathsf{ZF}$, choice is equivalent to the determinacy of all closed games on all sets."
EDIT: I've been asked to clarify the claim that $\mathsf{ZF}$ proves Borel determinacy (over $\omega^\omega$). This is indeed a bit tricky; the point is that we can use absoluteness theorems, and a bit of creative sentence rewriting, to remove any dependence on choice from the original proof (that said there's a subtlety around what "Borel" means without choice, see the very end):
Let $\mathcal{M}\models\mathsf{ZF}$. Now $\mathcal{M}$ may not have a forcing extension satisfying choice, but it does have one satisfying enough choice for the proof of Borel determinacy to go through; let $\mathcal{N}$ be such an extension. Now by Mostowski absoluteness, each Borel code in $\mathcal{M}$ is also a Borel code in $\mathcal{N}$ (the only nontrivial part of the definition of Borel code is well-foundedness). Let $b$ be a Borel code in the sense of $\mathcal{M}$, and let $b'$ be the thing which $\mathcal{M}$ thinks is the canonical Borel code for the complement of the set coded by $b$; by Mostowski as above, $b'$ is a code for the complement of $b$ in $\mathcal{N}$ as well. Then since $\mathcal{N}$ satisfies Borel determinacy, we have $\mathcal{N}\models$ "There is some strategy $\Sigma$ for (say) player 1 such that for every strategy $\Pi$ for player 2 and every real $F$ we have that $F$ is not a witness to $\Sigma\otimes \Pi$ being in the set coded by $b'$." Shifting to codes for complements has bought us a crucial quantifier: this latter statement is $\Sigma^1_2$ in a real parameter in $\mathcal{M}$ (namely $b$), and so by Shoenfield absoluteness holds in $\mathcal{M}$ as well as in $\mathcal{N}$.
This argument, though, reveals a subtlety which I had not mentioned originally: when we're working without choice, it's important to distinguish between two notions of "Borel"-ness. See this old answer of mine; to be absolutely precise, $\mathsf{ZF}$ proves Borel-coded determinacy for $\omega^\omega$.
