Let $\{X_{i}: i \in I\}$ be a family of $\sigma$-compact but not compact topological spaces. How to show that $\prod_{i \in I} X_{i}$ is $\sigma$-compact if and only if $I$ is finite?
So here it goes, using the hints by Sam and jspecter.
The direction $\Leftarrow$ indeed follows by induction, it suffices to do the case $n=2$. Indeed if $X_{1},X_{2}$ are $\sigma$ compact and $X_{1}=\cup C_{n}$, $X_{2}=\cup D_{n}$ now $X_{1} \times X_{2}$ = $\cup C_{n} \times \cup D_{n}$. Now here's my question: how do we write this? in general the cartesian product does not distributes with respect the union, although I guess that doesn't matter because iwe would get a countable union of finite product of compact sets, so its $\sigma$-compact.
For the other direction, suppose $I$ is infinite and suppose $\prod X_{i} = \cup_{n \in \mathbb{N}} C_{n}$ where each $C_{n}$ is a compact subset of $\prod_{i \in I} X_{i}$. Denote the projection by $p_{n}$ then $p_{n}(C_{n})$ is a compact subset of $X_{i}$. Since $X_{i}$ is not compact then $X_{i} \setminus p_{n}(C_{n})$ is non-empty, so for each $i \in I$ pick $z_{i} \in X_{i} \setminus p_{n}(C_{n})$, then $(z_{i}) \in \prod_{i \in I} X_{i}$, but not in the union $\cup C_{n}$, OK?