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I don't know enough logic to know if the reals possess a countably complete nonprincipal ultrafilter, but A has a winning strategy if the game is played on a set which has one. Choosing such an ultrafilter that contains the co-finite filter, A can always make their choice such that $\bigcap_{k\leq i} A_k$ is in the ultrafilter, since either $Q_i$ or $Q_i^c$ is in the ultrafilter and the filter is closed under finite intersections. Since it is countably complete, it is closed under countable intersections and so $\bigcap_{k \in \mathbb{N}} A_i$ lies in the ultrafilter as well. The intersection has infinitely many elements since the ultrafilter contains the co-finite filter, thus Q never does not win when A uses the strategy "pick whichever set is in the ultrafilter".

Edit: Thanks to those below who pointed out that $\mathbb{R}$ does not, in fact have a countably complete nonprincipal ultrafilter. This gives a cute proof of another partial result: Call a winning strategy for A strong if the choice of $A_i$ depends only on $Q_i$ and not on the previous state of the game. Then A has no strong winning strategy, for if A did, then the sets A would choose form a countably complete nonprincipal ultrafilter (the axioms are fairly easy to prove from the definition of the game).

If one could prove that a winning strategy for A implied the existence of a strong winning strategy for A, this would then show A has no winning strategy, but this seems beyond me.

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I don't know enough logic to know if the reals possess a countably complete nonprincipal ultrafilter, but A has a winning strategy if the game is played on a set which has one. Choosing such an ultrafilter that contains the co-finite filter, A can always make their choice such that $\bigcap_{k\leq i} A_k$ is in the ultrafilter, since either $Q_i$ or $Q_i^c$ is in the ultrafilter and the filter is closed under finite intersections. Since it is countably complete, it is closed under countable intersections and so $\bigcap_{k \in \mathbb{N}} A_i$ lies in the ultrafilter as well. The intersection has infinitely many elements since the ultrafilter contains the co-finite filter, thus Q never does not win when A uses the strategy "pick whichever set is in the ultrafilter".