Countable Product of Second-Countable Spaces is Second-Countable

Theorem
Let $\left \{{\left({S_\alpha, \tau_\alpha}\right)}\right\}$ be a countable set of topological spaces.

Let $\displaystyle \left({S, \tau}\right) = \prod \left({S_\alpha, \tau_\alpha}\right)$ be the product space of $\left \{{\left({S_\alpha, \tau_\alpha}\right)}\right\}$.

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

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

Proof
Let $X$ and $Y$ be second-countable spaces.

Let $\left\{{B_{x_i}}\right\}_{i \mathop \in \N}$ and $\left\{{B_{y_i}}\right\}_{i \mathop \in \N}$ be bases for their topologies.

By Cartesian Product of Countable Sets is Countable, the basis of $X \times Y$, namely $\left\{{B_{x_i} \times B_{y_j} }\right\}_{i, j \mathop \in \N}$, is also countable.

Let $\displaystyle \prod_{n \mathop \in \N} X_n$ be the product of topological spaces $\left({X_n, \tau_n}\right)$.

Let $B_n$ be a countable basis for $\tau_n$.

Then let $L_i = \left\{{\pi_i^{-1} \left({N_i}\right): N_i \in B_i}\right\}$.

Let $K_J = \displaystyle \bigcap_{j \mathop \in J} L_j$ for $J \subset \N$, $\left\vert{J}\right\vert < \infty$.

Then $\displaystyle B = \bigcup_{J \mathop \subset \N} K_J$ forms a basis of the product space.

Now, since the $K_J$'s can be identified with a finite product of countable sets, they are each countable

From Countable Union of Countable Sets is Countable, $B$ forms a countable basis of $\displaystyle \prod_{n \mathop \in \N} X_n$.