Given a topological space $\mathcal{X}=(X,\tau)$, the Banach-Mazur game on $\mathcal{X}$ is the (two-player, perfect information, length-$\omega$) game played as follows:
Players $1$ and $2$ alternately play decreasing nonempty open sets $A_1\supseteq B_1\supseteq A_2\supseteq B_2\supseteq ...$.
Player $1$ wins iff $\bigcap_{i\in\mathbb{N}} A_i=\emptyset$.
ZFC implies that there is a subspace of $\mathbb{R}$ with the usual topology whose Banach-Mazur game is undetermined; on the other hand, it's consistent with ZF+DC (and indeed adds no consistency strength!) that no subspace of $\mathbb{R}$ does this ("every set of reals has the Baire property").
However, when we leave $\mathbb{R}$ things get much weirder. My question is:
Does ZF alone prove that there is some space $\mathcal{X}$ whose Banach-Mazur game is undetermined?
Controlling the behavior of all possible topological spaces in a model of ZF is extremely hard for me, and I suspect the answer to the question is in fact yes. In fact, I recall seeing a pretty simple proof of this; however, I can't track it down or whip up a ZF-construction on my own (specifically, everything I try ultimately winds up being a recursive construction killed by having too many requirements to meet in the given number of steps).
This has now been partially answered at Mathoverflow by James Hanson, with the remaining case separately asked; I'm posting this answer to move this question off the unanswered queue (and I've made it CW so I don't get reputation for his work). The answer is affirmative under DC, and currently wide open otherwise.