Given a topological space $X$, Let $\mathcal O$, denote the set of all open covers of $X$.
We say that a space $X$ satisfies $S_1(\mathcal O,\mathcal O)$, if for every sequence of open covers $\{ \mathcal U_n : n \in \mathbb N \}$, there exists a sequence of open sets $\{ U_n : U_n \in \mat...
In the paper linked in the question the following claim is mentioned as Theorem 1: $X$ has $S_1(\mathcal O,\mathcal O)$ if and only if ONE does not have a winning strategy in $G_1(\mathcal O,\mathcal O)$.
Asking whether this is equivalent to: "two has a winning strategy in $G_1(\mathcal O,\mathcal O)$" is the same as asking whether every space fulfilling $S_1(\mathcal O,\mathcal O)$ is determined.
(If I'm not mistaken, the word determined is used in this context in the meaning that at least one of the players has winning strategy.)
If I understand correctly Corollary 2 in this paper, it says that if $X$ is a Lusin set, then this game is undetermined. Also the proof mentions that ONE does not have winning strategy, so it should be an $S_1(\mathcal O,\mathcal O)$ space.
Given a topological space $X$, Let $\mathcal O$, denote the set of all open covers of $X$.
We say that a space $X$ satisfies $S_1(\mathcal O,\mathcal O)$, if for every sequence of open covers $\{ \mathcal U_n : n \in \mathbb N \}$, there exists a sequence of open sets $\{ U_n : U_n \in \mat...
