Countable Product of Second-Countable Spaces is Second-Countable

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

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

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

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

Proof
First we show the finite case: Let $X$ and $Y$ be second-countable spaces and let $\{ B_{x_i} \}_{i \in \mathbf{N}}$, $\{ B_{y_i} \}_{i \in \mathbf{N}}$ be bases of their topologies. By Cartesian Product of Countable Sets is Countable the basis of $X \times Y$, namely $\{ B_{x_i} \times B_{y_j} \}_{i, j \in \mathbf{N}}$ is also countable.

Let $\prod_{n \in \mathbf{N}} X_n$ be the product of topological spaces $(X_n, \tau_n)$ and $B_n$ a countable basis for $\tau_n$. Then let $L_i = \{ \pi_i^{-1} (N_i) : N_i \in B_i \}$. Let $K_J = \bigcap_{j \in J} L_i$ for $J \subset \mathbf{N}$, $\vert J \vert < \infty$. Then $B = \bigcup_{J \subset \mathbf{N}} K_J$ 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. A countable union of countable sets (there are only countably many finite subsets in a countable set) is again countable by the axiom of choice. Hence, $B$ forms a countable basis of $\prod_{n \in \mathbf{N}} X_n$. $\square$