# Let $\tau$ denote the product topology on $X=X_1 \times X_2$. Then $(X,\tau)$ is a separable space.

Let $$(X_1,\tau_1)$$ and $$(X_2,\tau_2)$$ be separable spaces, and let $$\tau$$ denote the product topology on $$X=X_1 \times X_2$$. Then $$(X,\tau)$$ is a separable space.

Proof. $$(X_1,\tau_1)$$ and $$(X_2,\tau_2)$$ be separable spaces. So, there exists countable dense subsets $$D_1$$ of $$(X_1,\tau_1)$$ and $$D_2$$ of $$(X_2,\tau_2)$$ such that $$\overline {D_1}=X_1$$ and $$\overline {D_2}=X_2$$. Let $$(x_1,y_1)\in X_1 \times X_2 \implies x_1 \in X_1$$ and $$x_2 \in X_2$$. Let $$B\in \tau$$ such that $$(x_1,x_2)\in B$$. So, there exists Let $$U_1 \in \tau_1$$ such that $$x_1 \in U_1$$. $$U_1\cap D_1 \neq \phi$$. Let $$y_1 \in U_1\cap D_1$$. similarly, Let $$U_2 \in \tau_2$$ such that $$x_2 \in U_2$$. $$U_2\cap D_2 \neq \phi$$. Let $$y_2 \in U_2\cap D_2$$. So, $$(x_1,x_2) \in \overline {D_1 \times D_2}$$. We know that $$D_1 \times D_2$$ countable. Hence $$(X,\tau)$$ is separable. Am I correct?

Let $$W$$ be non-empty open subset in $$X_1×X_2$$, choose an element $$(x_1,x_2)\in W$$. Now by definition of basis of product topology we have an open set $$U\subseteq X_1$$ containing $$x_1$$ and an open set $$V\subseteq X_2$$ containing $$x_2$$ such that $$U×V\subseteq W$$.
Now $$\overline {D_1}=X_1$$ and $$\overline {D_2}=X_2$$ so that $$U\cap D_1\not=\phi,V\cap D_2\not=\phi$$ . So that $$B\cap (D_1×D_2)\supseteq (U×V) \cap (D_1×D_2)=(U\cap D_1)×(V\cap D_2)\not=\phi$$Therefore $$D_1×D_2$$ is dense in $$X_1×X_2$$.