TITLE: Game involving 'asking questions about a real'
QUESTION [26 upvotes]: Consider the following game, played by two players, 
called Q and A, in a time frame t = 1, 2, ....
At every time point i, Q mentions some $Q_i \subset \mathbb{R}$,
after which A mentions $A_i$ such that either $A_i = Q_i$ or
$A_i = Q_i^c$.
Define $C(i) = \bigcap_{k \lt i} A_i$ and
$C(\infty) = \bigcap_{k \in \mathbb{N}} A_i$.
Player Q is declared winner if either:
1) $C(i)$ has only one element for some $i < \infty$
or
2) $C(\infty)$ is empty.
(Player A wins in all other cases.)
Question: 
Which player (if any) has a winning strategy?
Edit:
Some additional explanatory remarks:
This question occurred to me while thinking about Brouwer style choice sequences in connection with Borel sets.
Before posting, I was not 100% sure yet, but pretty confident that player A has a winning strategy when Q is restricted to giving Borel sets. This has been confirmed, in the meantime, in one of the answers below. (More generally A has a winning strategy when Q is restricted to sets having the Baire property.)
The general case is still open, though.
Some easy observations that might help to clarify the puzzle:
a) Player Q obviously has a strategy to make sure that $C(\infty)$ has at most one element. But that's not enough to win the game. 
b) Whenever $C(i)$ is countable for some (finite) $i$, player Q can win the game, by (from time $i$ onwards) eliminating the elements of that countable set, one by one.
c) Player A has an easy strategy to make sure that $C(i)$ is uncountable for all (finite) $i$. But that's not enough to describe a winning strategy for player A. For example, when $Q_i = A_i = (0, 1/i)$ for every $i$, then $C(i)$ is uncountable for every (finite) $i$, but $C(\infty)$ is empty, and therefore A loses the game.

REPLY [2 votes]: 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.