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-cooridinate 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.