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For the opposite direction, given that X and Y are compact we start with taking $O$ to be an open cover for $X \times Y$. Then for each $x \in X$, ${x} \times Y$ is compact. Later to conclude that the union of open sets $O_x$ is open and for each $x \in X$ there exists an open $W_x$ $\subset X$ such that $W_x \times Y$ is covered by $O_x$. Along those lines... but how do I go backwards?

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The answer below gives a more elegant solution, but starting with an open cover of $X$, what is the only open cover of $X\times Y$ that comes to mind that can be created from it? – Carsten S Nov 4 '13 at 11:13

The projection map into the individual spaces/components is a continuous map, so that it sends compact to compact.

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And one sees nicely that an assumption is missing in the claim, which is wrong as stated. – Carsten S Nov 4 '13 at 11:11
@CarstenSchultz: For each space to be the image of the projection map, all spaces must be non-empty. Is that what you mean? – Stefan Hamcke Nov 4 '13 at 15:20
@StefanH Exactly. – Carsten S Nov 4 '13 at 15:22

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