Consider the following infinite game: two players, I and II, are alternating and choosing a descending sequence of subsets of $\mathbb R$ of cardinality $\frak c$, so I chooses a set $A_1\subseteq\mathbb R$, II chooses a set $A_2\subseteq A_1$, I chooses $A_3\subseteq A_2$ etc., all having the size continuum. Let $A=\cap_{i=1}^\infty A_i$. I wins if $A$ is nonempty, otherwise II wins.

Who has a winning strategy in this game?

This game was posed to me as a puzzle from a person I don't have any means of contact with anymore (so I can't ask them about this). However, I don't know whether he himself knew the answer or not (as far as I know it might be independent of ZFC or unknown) which is why I'm posting this on MO and not Math.SE (the same person has posed as a puzzle a combinatorial problem which I happened to know is open).

All I know is that it's consistent with ZF that II has the winning strategy: if $\mathbb R$ had countable cofinality, i.e. was a countable union of sets $S_i,i\in\mathbb N$ of cardinality $<\frak c$, then II could play on his turns $A_{2k}=A_{2k-1}\setminus(S_1\cup\dots\cup S_n)$, which would have size $\frak c$. This doesn't go through in ZFC, since there $\frak c$ has uncountable cofinality.

One can also consider the problem with $\mathbb R$ replaced by some other set, for example an ordinal. For some sufficiently well-behaved sets (not Dedekind-finite, for example) the above reasoning shows that if a cardinal has countable cofinality, I has a winning strategy. I am especially interested in the case of $\omega_1$, which is equivalent to the problem at hand if we assume CH.

I am interested in this problem both in context of ZF and ZFC, and I am willing to assume CH to resolve the original problem if it is of help there.