Choosing subsets of $\mathbb R$ of cardinality $\frak c$, who wins?

In ZFC, the player aiming for the empty set has a winning strategy in the game played on any infinite set, including the reals. Using the axiom of choice, we can well-order the set and thereby pretend that we are playing sets of ordinals. Now, whenever it is the empty-set player's turn, she should look at the order type of the ordinals remaining, and play so as to omit the first element of every successive block of $\omega$ many elements in the enumeration of the set. In this way, no element can survive into the final intersection. If an element is the $(k+1)^{th}$ in it's $\omega$-block, then it is the $(\omega\cdot\gamma+k+1)^{th}$ element of $A_n$ for some ordinal $\gamma$, and after the empty-set player plays, it will become the $(\omega\cdot\gamma+k)^{th}$ element of the new set, strictly smaller. When the other player plays, if it was the $\alpha^{th}$ element, it becomes the $\beta^{th}$ element for some $\beta\leq\alpha$. So as play progresses, the index of the element in the enumeration will steadily descend. Since it cannot descend forever, the element must eventually fall out of the set and so the final intersection is empty, as desired.

In ZF, the argument I gave at the other question shows that the non-empty player cannot have a winning strategy in the game on the reals.

This problem (with inconsequential differences in the rules of play) was posed by S. Banach as Problem 67 in The Scottish Book, which is available here. The following text is copied from R. Daniel Mauldin's edition:

(A MODIFICATION OF MAZUR'S game, [see Problem 43]).

We call a half of the set $E$ [in symbols, $(1/2)E$] an arbitrary subset $H\subset E$ such that the sets $E,H,E-H$ are of equal power.

(1) Two players $A$ and $B$ give in turn sets $E_i, i=1,2,\dots$ ad inf. so that $E_i=(1/2)E_{i-1}\ i=1,2,\dots$ where $E_0$ is a given abstract set. Player $A$ wins if the product $E_1E_2\dots E_iE_{i+1}\dots$ is vacuous.

(2) The game, similar to the one above, with the assumption that $E_i=1/2[E_0-E_1-\dots-E_{i-1}]\ i=2,3,\dots$ ad inf., and $E_1=(1/2)E_0.$ Player $A$ wins if $E_1+E_2+\dots=E_0.$

Is there a method of win for player $A$? If $E_0$ is of power cofinal with $\aleph_0,$ then player $A$ has a method of win. Is it only in this case? In particular, solve the problem if $E_0$ is the set of real numbers.

Addendum. There exists a method of play which will guarantee that the product of the sets is not vacuous. The solution was given by J. Schreier.

The problem is dated August 1, 1935; the addendum is dated August 24, 1935; Schreier's solution (using the well-ordering theorem, see Joel David Hamkins' answer) was published in the paper

J. Schreier, Eine Eigenschaft abstrakter Mengen, Studia Math. 7 (1938), 155–156, eudml.