It is a very nice question.
I claim that it is not sufficient that $C$ is determined, and indeed, there are counterexamples where $C$ is a game with only two moves.
Consider the two-dimensional game where player I plays $(x_0,y_0)$ and player II responds with $(x_1,y_1)$. Player I wins, if $y_1=x_0$. That is, player I wins, if player II copies on his second coordinate the first-coordinate move of player I. In your framework, the payoff set $C$ is the set of plays with projections to $(x,y)$, where $y(1)=x(0)$.
Clearly, player II has a winning strategy in this game, which is simply to make sure that $y_1$ is not the same as the already-played $x_0$.
But I claim that there can be no strategies $\sigma$ and $\tau$ for player I with $[\sigma]\times[\tau]\subset C$ or for player II, with $[\sigma]\times[\tau]\subset\neg C$.
In the first case, for any one-dimensional strategies $\sigma$ and $\tau$ for player I, we can devise a play that refutes them by having player II actually play so as to violate the move-copying requirement.
In the second case, for any one-dimensional strategies $\sigma$ and $\tau$ for player II, we can have player I first play $y_0$, in order to get $\tau$'s response, and then play $x_0$ using that information. In this way, player I can in effect look ahead in the second coordinate to see how player II will play, and then using that information complete the first move in the first coordinate by playing $x_0$ in such a way that player II will in effect have copied it. So this play will be in $C$, contrary to hypothesis.
So there are counterexamples with clopen games of very low complexity.