Infinite Product of Sigma-Compact Spaces is not always Sigma-Compact

Theorem
Let $I$ be an indexing set with infinite cardinality.

Let $\left\langle{\left({S_\alpha, \tau_\alpha}\right)}\right \rangle_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.

Let $\displaystyle \left({S, \tau}\right) = \prod_{\alpha \mathop \in I} \left({S_\alpha, \tau_\alpha}\right)$ be the product space of $\left\langle{\left({S_\alpha, \tau_\alpha}\right)}\right \rangle_{\alpha \mathop \in I}$.

Let each of $\left({S_\alpha, \tau_\alpha}\right)$ be $\sigma$-compact.

Then it is not necessarily the case that $\left({S, \tau}\right)$ is also $\sigma$-compact.

Proof
Let $T = \left({\Z_{\ge 0}, \tau}\right)$ be the topological space formed by the discrete topology on the set of positive integers.

Let $T' = \left({\displaystyle \prod_{\alpha \mathop \in \Z_{\ge 0} } \left({\Z_{\ge 0}, \tau}\right)_\alpha, \tau'}\right)$ be the countable Cartesian product of $\left({\Z_{\ge 0}, \tau}\right)$ indexed by $\Z_{\ge 0}$ with the Tychonoff topology $\tau'$.

From Countable Discrete Space is Sigma-Compact, $T$ is $\sigma$-compact.

From Countable Product of Countable Discrete Spaces is not Sigma-Compact, $T'$ is not $\sigma$-compact.

Hence the result.