Finite Product of Sigma-Compact Spaces is Sigma-Compact

Theorem
Let $n \in \Z_+^*$ be a positive integer.

Let $\left \{{\left({S_i, \tau_i}\right): 1 \le i \le n}\right\}$ be a finite set of topological spaces.

Let $\displaystyle \left({S, \tau}\right) = \prod_{i=1}^n \left({S_i, \tau_i}\right)$ be the product space of $\left \{{\left({S_i, \tau_i}\right): 1 \le i \le n}\right\}$.

Let each of $\left({S_i, \tau_i}\right)$ be $\sigma$-compact.

Then $\left({S, \tau}\right)$ is also $\sigma$-compact.

Proof
Supose that $\left({S_1, \tau_1}\right)$ and $\left({S_2, \tau_2}\right)$ are $\sigma$-compact.

Then $\displaystyle S_1 = \bigcup_{n=1}^\infty C_n$ and $\displaystyle S_2 = \bigcup_{n=1}^\infty D_n$ where all the $C_n$ and $D_n$ are compact sets.

From Tychonoff's Theorem, the set $\mathcal K = \{C_n \times D_m: n, m \in \N\}$ is a countable set of compact sets of $S_1 \times S_2$ with the product topology.

Consider $(a,b)\in S_1 \times S_2$.

Then $a \in S_1$ and $b \in S_2$.

Using the $\sigma$-compactness of $S_1$ and $S_2$, $\exists n, m$ such that $a \in C_n$ and $b \in D_m$.

So $(a,b)\in C_n \times D_m \in \mathcal K$.

Thus, $\bigcup \mathcal K = S_1 \times S_2$ and then $S_1 \times S_2$ is $\sigma$-compact.

Using induction on the number of factors of the product we get the result.