Studying the article "Games that involve set theory or topology" of Marion Scheepers, I found the following result

**Theorem 46** Let $\{(X_{i}, \tau_{i}) : i\in I  \}$ be a family of topological spaces. If for every countable subset $J$ of $I$, Player I does not have a winning strategy in the game $\textsf{BM} (\prod_{j\in J}X_{j},
			\square_{j\in J} \tau_{j}  ),$ then Player I does not have a winning strategy in $\textsf{BM} (\prod_{i\in I}X_{i},
			\prod_{i\in I} \tau_{i}  ).$

Where $\textsf{BM}(X)$ denotes the Banach-Mazur game played on a topological space $X$. 

So far I had no success in proving the theorem, the author only states the result without mentioning any reference. I would like to know if you could give me some idea about it.

Remember that The Banach-Mazur game on $X$, $\textsf{BM}(X)$, is played as follows: Players I
and II play an inning per positive integer. In the $n$-th inning Player I chooses a nonempty open set $A_n$; Player II responds with a nonempty open set $B_n \subseteq A_n$. Player I must also obey the rule that for each $n$, $A_{n+1} \subseteq B_n$. 
A play $A_1, B_1, \cdots A_n, B_n$ is won by Player II if $\bigcap_{n\in\omega}T_n \not=\emptyset$; otherwise, Player I wins.

One of the most important results in this regard is the following
 
**Theorem**  A nonempty topological space $X$ is a Baire space if and only if Player I has no winning strategy in the Banach-Mazur game $\textsf{BM}(X)$. 



Thanks