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**Defunkt** I think I came up with a solution. Tell me if you see anything wrong please.

I want to show that f is continuous, ie. that if $\displaystyle V \subset Y$ is open then $\displaystyle f^{-1}(V)$ is open as well.

Let V be an open subset of Y. Let $\displaystyle U = V^c$, then U is closed, and so is $\displaystyle X \times U$. Since $\displaystyle G$ is closed as well, $\displaystyle G \cap (X \times U)$ is also closed.

From a previous result, we know that the projection $\displaystyle \pi_x:X \times Y \to X$ is a closed mapping when Y is compact. Therefore, $\displaystyle \pi_x(G \cap (X \times U) )$ is closed.

However, note that $\displaystyle \pi_x(G \cap (X \times U) )$ is exactly the set of points in X which f does not map into V, and it is closed, therefore $\displaystyle f^{-1}(V)$ is open and we are done.

I know it is a bit informal, but that is simply the outline.

Thanks again for helping.