Step in proof that product of compact spaces is compact

Question

Many proofs of the fact that the product of compact spaces use the tube lemma, and therefore start somewhat like this: Let $X$ and $Y$ be compact spaces. Then for each $x\in X$, $\{x\}\times Y$ can be covered by finitely many open sets. Why is this true? I understand intuitively but am having a hard time rigorously justifying this.

Answer 1

Consider how the topology of $\{x\}\times Y$ is defined, and you will see that $f(p)=(x,p)$, for $p\in Y,$ is a homeomorphism from $Y$ to $\{x\}\times Y.$ Compactness is a topological property, that is, it is preserved by homeomorphism. And $Y$ is compact, so $\{x\}\times Y$ is compact.

Answer 2

The subspace $\{x\}\times Y$ is homeomorphic to $Y$, which is compact by assumption. Compactness is a topological property, meaning if $X\cong Y$ and $X$ is compact, then $Y$ is compact as well.