Hi bof, just to let you know, I recently proved a few theorems that improve on the work of Bankston and McGovern that you mention in your second paragraph. It turns out their theorems are provable in ZFC (you just have to work a bit harder without MA or CH), and in fact can be strengthened as well. The paper is here: sciencedirect.com/science/article/abs/pii/S0166864122002826. The main results are: (1) If a metric space has size $\leq \mathfrak{c}^{+\omega}$, then it can be partitioned into copies of the Cantor space if and only if it can be covered with copies of the Cantor space.

@StevenClontz: Yes, you're right. Also, I posted my comment before taking the time to digest Joel's. I think a simple modification of Debs' space cannot answer Monroe's question, unless the modification somehow avoids the argument given in bof's answer.

Steven, what if you use the poset on $\omega_1 \times (X \cup \mathcal T_X^{\neq \emptyset})$ where $(\alpha,U) \leq (\beta,V)$ if and only if $\alpha > \beta$ and $U \subseteq V$, and $(\alpha,x) \leq (\beta,V)$ if and only if $\alpha > \beta$ and $x \in V$, and $(\alpha,x) \leq (\beta,y)$ if and only if $\alpha > \beta$ and $x=y$?

@JoelDavidHamkins: Debs' space does indeed have bof's property. This is essentially how he proves that II has a winning $2$-tactic in the BM game.

. . . but it does not answer the poset version. In fact, one of Debs' spaces (there are two -- one that's Hausdorff but non-regular, and a much more complicated one that is $T_{3 \frac{1}{2}}$) refines a metric space, and this means that player I can play in such a way that the sequence of moves is guaranteed to contain at most one point in the end, and therefore not contain a nonempty open set.

Hi Joel, unfortunately, I don't think Debs' example answers Monroe's question. The problem is that the Banach-Mazur game on a topological space is not generally equivalent to the Banach-Mazur game on the poset consisting of its nonempty open sets. The problem is that in the topological version, II wins if there is a point contained in every set that's been played. In the poset version, II wins if there is a member of the poset included in every set that's been played. These are different conditions. Debs' space answers the topological-Banach-Mazur version of this question . . . (continued)

@AsafKaragila: I've been using MO for over 7 years and somehow never noticed :) Thanks for letting me know!

@Wojowu: That looks right. I wasn't thinking very carefully, and I remember that in the topological category, $\omega^*$ is not homeomorphic to its square. I was guessing this should imply $\mathcal P(\omega)/\mathrm{fin}$ is not isomorphic to its square (as posets), but I suppose that's not so.

@Wojowu: You sure about that?

This is an interesting question. I feel like $I = \omega$ and $X = (2^\omega)^\omega_{box}$ should work, but I don't see how to prove it.

Hi Federico, I've written an answer to your question, but it doesn't seem to be displaying properly on the MO homepage. I'm not sure what's wrong, or whether it's an issue with my computer or with MO. But I hope the answer is visible to you?