A question about a two player game and axiom of choice Suppose two players 1 and 2 play the following game:
Player 1 starts by playing the set of reals $\mathbb{R}$. Player 2 plays an uncountable subset $Y_1$ of $\mathbb{R}$. Then player 1 plays an uncountable subset $X_1$ of $Y_1$. Then player 2 plays an uncountable subset $Y_2$ of $X_1$ and so on. Player 1 wins if the intersection of all the sets played is non empty otherwise player 2 wins.
Then one can show that if $\omega_1$ injects into $\mathbb{R}$ then player 2 has a winning 
strategy.
Question 1: Can the existence of a winning strategy for player 2 be shown in $ZF + DC$?
Question 2: If not, then is the following consistent: $ZF + DC +$ "Player 1 has a winning strategy"? A possible model for this could be a model of "Every uncountable set of reals has a perfect subset" but I don't know anymore.
 A: I think that the following is a correct proof of the following claim:

Proposition ($\mathsf{ZF+DC}$). If Player I has a winning strategy then $\omega_1\leq\mathbb R$.

Proof. Let $F\colon\mathbb R\to\omega_1$ be a surjection such that each fiber is uncountable (which is definable in $\mathsf{ZF}$), and let $q_n$ be an enumeration of the rationals.
Let $\alpha<\omega_1$, we consider the following game: $Y_1=F^{-1}(\alpha)$. This is an uncountable set, so it is a legal move. Player I plays by its strategy, and Player II plays by the following strategy:
Suppose $X_n$ was chosen, let $q$ be the least rational number such that $(q-\frac1n,q+\frac1n)\cap X_n$ is a legal move. Such rational exists otherwise $X_n$ is the countable union of countable sets, which under $\mathsf{DC}$ is countable. Then $Y_n$ is that intersection.
As the first player has a winning strategy it assures us that $\bigcap Y_n\neq\varnothing$, but in this case the intersection can only contain one point, because $a,b\in Y_n$ for all $n$ it means that $|a-b|<\frac1n$ for all $n$. We denote $r_\alpha$ this unique point.
It is left to show that $f(\alpha)=r_\alpha$ is injective, but this is trivial because $F(r_\alpha)=\alpha$ by the fact that $r_\alpha\in F^{-1}(\alpha)$. $\square$

From this follows that it is impossible in $\mathsf{ZF+DC}$ to have a winning strategy for Player I. But it is not enough to prove the existence of a strategy for Player II.
I think that the only use of the axiom of choice is in showing that countable unions of countable sets of real numbers are countable. If this is false, Andres gave a strategy for Player I, and if this is true the arguments above along with the arguments in the comments to the question show that Player I cannot have a winning strategy. Indeed we are left to show whether or not $\mathsf{ZF+DC}$ prove that Player II can win, or maybe there is some model in which the game is indeterminate.
A: Let me first point out that (in $\mathsf{ZF}$) if there is a countable collection of countable sets of reals whose union is uncountable, say $Y=\bigcup_n A_n$, where we may as well assume the $A_n$ are disjoint, then II has a winning strategy: For each $n$, II wins by playing $Y_{n+1}=X_n\cap\bigcup_{m>n}A_m$, where $X_0=\mathbb R$. Note each $Y_{n+1}$ has countable complement in $X_n$, and their intersection is contained in $\bigcap_n\bigcup_{m>n}A_m=\emptyset$, so this is winning for II. 
An earlier version of this answer had a mistake. Asaf has given the right answer, let me add some remarks: Asaf proof actually shows that, if every countable union of countable sets of reals is countable, and I has a winning strategy, then $\omega_1\le \mathbb R$. As described in the comments, if $\omega_1\le\mathbb R$, then I was a winning strategy. It remains to address what happens when $\omega_1\not\le\mathbb R$ but countable unions of countable sets of reals are countable. (The two possibilities are that the game is undetermined, or that II has a winning strategy.)
