TITLE: A question about a two player game and axiom of choice
QUESTION [15 upvotes]: 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.

REPLY [3 votes]: 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.)