@caffeinemachine: I don't have a reference. But I would say it's really just a corollary to the Hales-Jewett Theorem.

What if we take $X(\alpha) = Ord$ for all $\alpha \geq \omega$, and define $X(n) = \{ \beta :\, (V_\beta,\in) \vDash \phi_n \}$, where $\phi_0,\phi_1,\phi_2,\dots$ is some enumeration of the ZFC axioms? ZFC cannot prove that the diagonal intersection is nonempty here, much less a club, but each $X(n)$ is a club by the reflection principle. And I think this is a definable class of the form you're asking about. (But I'm not certain I haven't stepped on some Gödelian landmine with this last assertion, which is why I'm posting this as a comment rather than an answer.)

@Agelos: Yes, I think you're right. A clever example!

Also nontrivial: If $A$ and $B$ are two closed, countable, nonhomeomorphic subspaces of $\mathbb R^3$, does this imply that $\mathbb R^3 \setminus A$ and $\mathbb R^3 \setminus B$ are not homeomorphic?

Correction to my first comment: if player 1 plays $k_0 = 1$ then player 2 can respond with $k_1 = 2$ or $3$. But it is true that if either player plays $k_{n+1} = k_n+1$ for any $n \geq 0$ (any time after the very first move), then this forces $k_{m+1} = k_m+1$ for all $m > n$ and effectively ends the game.

I don't know about annoying -- I think it makes things interesting. My first thought was that your game should be pretty similar to one called Baker's game, where players alternately choose $a_0,b_0,a_1,b_1,\dots$ in such a way that $a_0 < a_1 < \dots < b_1 < b_0$, and then player 1 wins if $\lim a_n$ is in some payoff set. This game is well understood: player 1 wins if and only if the payoff set contains a perfect set, and player 2 wins if and only if the payoff set is countable. But after thinking about it for a bit, your game seems much weirder than this one. (In a good way!)

One strange-seeming observation: if player 1 plays $k_0=1$, then the players must form the alternating harmonic series, and player 1 wins if and only if $\ln(2) \in A$. Similarly, either player has the freedom, at any point in the game, to effectively end things by playing $k_{n+1} = k_n+1$, as your rules then force $k_{m+1} = k_m+1$ for all $m > n$ as well.

I don't know the answer to this question, but I think this thread is relevant: mathoverflow.net/questions/324254/…. If I understand correctly, Andreas' answer there blocks a plausible way to proving a positive answer to your question.

But I think your number is still $\mathfrak{r}$. If $\mathcal R$ is a reaping family, then closing $\mathcal R$ under finite modifications gives a non-bitartite family. And if $\mathcal E$ is a non-bipartite family, then it is also a reaping family.

@DominicvanderZypen: You're right! I missed that.