A two player game on compact topological spaces I've though of an infinite game that two players may play on a given topological space $(X,\tau)$. It goes like this. 
On turn $n$ Player I selects a point $x_n\in X$ and Player II selects a neighborhood $U_n$ of $x$. 
They play $\omega$ times, and Player I wins if $\{U_n\}_n$ is a cover of $X$. 
Thinking about it I've come up with what I believe are some true statements regarding this game. 
-If $(X,\tau)$ is not Lindelöf then Player II has always a winning strategy.
-If $(X,\tau)$ is countable than Player I has always a winning strategy.
-If $X=[0,1]$ then Player II has a winning strategy (just choose interval neighborhoods with increasingly smaller diameter, in such a way that the series given by adding them converges to a value smaller that $1$).
-If $X=\omega_1+1$ (or any successor ordinal) then Player I has a winning strategy. 
Now I think it might be that, given that $(X,\tau)$ is compact, one of the players always has a winning strategy (i.e. the game is determined), because open games are determined. I mean, this depends on whether open games on game spaces of the form $S^\omega$, not necessarily the Baire space, are always determined. 

My question is whether it is known if this dividing line among compact topologies, namely whether Player I or Player II has a winning strategy in this game, has strong consequences in any direction. 

For example it might be that this can be framed in terms of classical topological properties. 
 A: According to the Wikipedia article on topological games, your game is known as the point-open game. Wikipedia has no more information about this game, but refers to the paper Rastislav Telgârsky, Topological games: on the 50th anniversary of the Banach–Mazur game, Rocky Mountain J. Math. 17 (1987), 227–276, which has a lot of information about it.
A: Neat question, and it has a neat answer for compact Hausdorff spaces.  Namely, Player II can win in a compact Hausdorff space $X$ iff $X$ contains a nonempty perfect subset (a closed subset with no isolated points).
First, suppose $X$ has a nonempty perfect subset $A$.  We may as well assume $X=A$ then (if Player I picks a point outside $A$, just pick $X\setminus A$ as the open set), so that $X$ is perfect.  Now note that if $U\subseteq X$ is open, then $U$ has no isolated points either and so $\overline{U}$ is perfect as well.  So, when Player I picks a point $x_1$, here's what Player II can do.  Since $x_1$ is not isolated, there exists another point $y\in X$.  Since $X$ is Hausdorff, there are disjoint open sets $U,V\subset X$ with $x_1\in V$ and $y\in U$.  Player II then chooses $U_1=X\setminus \overline{U}$.
Now $A=\overline{U}$ is again a nonempty perfect subset of $X$, and so we may again replace $X$ with this subset for subsequent turns.  So, Player II can continue the game for any number of turns, ensuring that after every turn, there is still a nonempty perfect set which has not been covered by the open sets chosen so far.  By compactness, the open sets cannot cover the whole space after infinitely many steps either, so Player II wins.
Conversely, suppose $X$ has no nonempty perfect subset.  Recall that the Cantor-Bendixson derivative $X'$ of $X$ is the complement of the set of isolated points of $X$.  We can then iterate this process transfinitely, defining $$X_0=X,$$ $$X_{\alpha+1}=X_\alpha',$$ and $$X_\alpha=\bigcap_{\beta<\alpha}X_\beta$$ when $\alpha$ is a limit ordinal.  Each $X_\alpha$ is a closed subset of $X$, and $X_{\alpha+1}$ is always a proper subset of $X_\alpha$ (unless $X_\alpha$ is empty) since $X_\alpha$ is not perfect.  So, there must exist some ordinal $\alpha$ such that $X_\alpha=\emptyset$.  The least such $\alpha$ will be a successor, and its predecessor $\beta$ is called the Cantor-Bendixson rank of $X$.
We now prove that Player I has a winning strategy, by induction on the Cantor-Bendixson rank $\beta$.  Note that since $X_{\beta+1}=\emptyset$, $X_\beta$ is a discrete space, which must be finite by compactness.  So, Player I uses their first $|X_\beta|$ turns to pick each of the points of $X_\beta$.  After those turns, let $Y\subset X$ be the subspace that has still not been covered by the open sets chosen.  It is clear by induction that $Y_\alpha\subseteq X_\alpha$ for all $\alpha$, and so $Y_\beta=\emptyset$ since $Y$ contains no points of $X_\beta$.  Thus $Y$ has Cantor-Bendixson rank less than $\beta$.  So, by induction, Player I can win the game on $Y$.  Since they have already covered all the points of $X\setminus Y$, this is enough to win the game on $X$.
(In fact, we can relax the assumption that $X$ is compact Hausdorff a bit.  For Player II to win by the first argument, we need only for $X$ to contain a nonempty closed compact Hausdorff subset with no isolated points.  And for Player I to win by the second argument, we only need $X$ to be compact and contain no nonempty perfect subset.)
