= Solution
For a finite position $s\in A^{<\omega}$, let $[s]$ be the <cylinder set> of infinite plays extending $s$. Because the factors have the <discrete topology>, these <cylinder sets> form a basis for the <product topology>. Put
$$
D=\{s:[s]\subseteq G\}.
$$
Membership in $D$ means that I has already secured the outcome, regardless of future moves. Since $G$ is an <open set>, a play belongs to $G$ exactly when one of its prefixes belongs to $D$.
Build the <winning-position attractor in an infinite game> by <transfinite recursion>. Start with $W_0=D$, use unions at limit <ordinals>, and set
$$
W_{\alpha+1}=D
\cup\{s:|s|\text{ is even and }\exists a\in A\ (sa\in W_\alpha)\}
\cup\{s:|s|\text{ is odd and }\forall a\in A\ (sa\in W_\alpha)\}.
$$
This sequence is increasing: the defining operation is <order-preserving>, and $W_0\subseteq W_1$. It stabilizes at some ordinal. Indeed, if it strictly increased at every successor stage below the successor <cardinal number> of $|A^{<\omega}|$, choosing one newly added position per stage would inject that larger cardinal into $A^{<\omega}$. Denote the stable set by $W$. This constructs the needed <fixed point> directly, without an appeal to an unstated <fixed-point theorem>.
Every position in $W$ has a least entry <ordinal>, its rank. Rank zero means membership in $D$. A positive rank is a successor $\alpha+1$: at an I-position some extension belongs to $W_\alpha$, while at a II-position every extension does. If the <empty word> belongs to $W$, I always chooses an extension of smaller rank until $D$ is reached. II's choices also decrease rank before that happens. There is no infinite strictly decreasing sequence of <ordinals>, so some finite prefix lies in $D$. This is a <winning strategy in an infinite game> for I.
If the <empty word> is outside $W$, the <fixed point> equation says that every extension of an I-position outside $W$ is still outside $W$, and that a II-position outside $W$ has at least one extension outside $W$. II chooses such an extension. Every resulting prefix avoids $D$, hence the whole play avoids $G$ by openness. This is a <winning strategy in an infinite game> for II.
The choices can be made simultaneously by fixing a <well-order> of $A$, as allowed by the usual <axiom of choice>. The argument does not assume that $A$ is countable. \b[Every such open game is determined]:
$$
\boxed{\varepsilon\in W\Rightarrow\text{I wins};\qquad
\varepsilon\notin W\Rightarrow\text{II wins}.}
$$
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