Proving the converse of "A Cartesian product of a finite number of compact spaces is itself compact."

Question

There is plenty of material online where it is shown that "A Cartesian product of a finite number of compact spaces is itself compact". I am wondering if the converse is true as well. That is

Let $\mathcal{X}\times\mathcal{Y}$ be a compact subspace. Does that imply that $\mathcal{X}$ and $\mathcal{Y}$ are compact as well?

I found a similar question here, but I am still not sure about it.

Answer 1

Hint: Let $\pi:X\times Y\to X$ be the projection onto the first coordinate. Then $\pi$ is continuous and $\pi[X\times Y]=X$.

Answer 2

Suppose $X_1$ is not compact and $\{X_i:i\in I\}$ is a family of non-empty spaces, with $1\not \in I\ne \emptyset.$ Let $C$ be an open cover of $X_1$ with no finite sub-cover. Then $\{c\times \prod_{i\in I}X_i: c\in C\}$ is an open cover of $X_1\times \prod_{i\in I}X_i$ with no finite sub-cover.