Uncountable Product of Second-Countable Spaces is not always Second-Countable

Theorem
Let $I$ be an indexing set with uncountable 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 a second-countable space.

Then it is not necessarily the case that $\left({S, \tau}\right)$ is also a second-countable space.

Proof
Let $T = \left({\Z_{\ge 0}, \tau}\right)$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology.

Let $I$ be an indexing set with uncountable cardinality.

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

From Countable Discrete Space is Second-Countable, $T$ is a second-countable space.

But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not Second-Countable, $T$ is not a second-countable space.

Hence the result.