The Wikipedia article on the Axiom of Determinacy (AD) claims:

Equivalent to the axiom of determinacy is the statement that for every subspace X of the real numbers, the Banach–Mazur game BM(X) is determined.

Is this claim true?

AD is usually stated for the Gale-Stewart game $G(S)$ with payoff set $S\subseteq \omega^\omega$, in which the players build a sequence in $\omega^\omega$ by alternately choosing integers, and Player I wins if this sequence is in $S$.

In the Banach-Mazur game $BM(X)$ with payoff set $X\subseteq \mathbb{R}$, the players build a descending sequence $U_0\supseteq U_1\supseteq U_2\supseteq \dots$ by alternately choosing nonempty open sets, and Player II wins if $\bigcap_{n\in \omega}U_n\subseteq X$.

AD implies the determinacy of $BM(X)$ for all $X\subseteq \mathbb{R}$, since $BM(X)$ can be replaced by an equivalent game in which the players are additionally required to play open intervals with rational endpoints, which can be coded by natural numbers.

After a bit of Googling, I couldn't find any information about the converse (aside from the claim on Wikipedia). Of course, this might just be because it's very obvious. But on the other hand, the fact that $BM(B)$ is determined when $B$ is Borel (or even just has the Baire property) is an easy theorem, while the corresponding fact for $G(B)$ is a very hard theorem ("Borel determinacy"). So I can believe that the converse might be false.

**Edit:** In light of the answer to this question, I've edited the Wikipedia page, replacing the (false) equivalence with the (true) implication:

The axiom of determinacy implies that for every subspace X of the real numbers, the Banach–Mazur game BM(X) is determined.