Countable Product of Separable Spaces is Separable

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

Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.

Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.

Let each of $\struct {S_\alpha, \tau_\alpha}$ be separable.

Then $\struct {S, \tau}$ is also separable.

Proof
For each $\alpha \in I$, let $D_\alpha$ be a countable everywhere dense subset of $\struct {S_\alpha, \tau_\alpha}$.

Let $D = \ds \prod_{\alpha \mathop \in I} D_\alpha$.

From Countable Union of Countable Sets is Countable, $D$ is a countable set of $S$.

From Natural Basis of Product Topology, the set $\BB$ of cartesian products of the form $\ds \prod_{\alpha \mathop \in I} U_\alpha$ where:
 * for all $\alpha \in I : U_\alpha \in \tau_\alpha$
 * for all but finitely many indices $\alpha : U_\alpha = X_\alpha$

is a basis for $\tau$.

From Analytic Basis Characterization of Denseness, to show that $D$ is an everywhere dense subset in $\struct {S, \tau}$ it is sufficient to show that:
 * $\forall U \in \BB : U \cap D \ne \O$

Let $U = \ds \prod_{\alpha \mathop \in I} U_\alpha \in \BB$.

Then:
 * $\forall \alpha \in I : U_\alpha \cap D_\alpha \ne \O$

For all $\alpha \in I$, let $x_\alpha \in U_\alpha \cap D_\alpha$.

Then:

Thus:
 * $U \cap D \ne \O$

The result follows.