Let $X_n$ be separable spaces for each $n$. Then, $X=\prod_nX_n$ is separable. (n is a natural number)
Note: I am considering the product topology. Note, $\textbf{x}=(x_1,x_2,x_3....)$
My attempt:
Let $D_n$ denote the countable dense subset of $X_n$. For each $k$, denote $\textbf{x}_k\in X$ to be such that $x_k\in D_k$ . Put $D=\bigcup_n \{ \textbf{x}_n: x_n\in D_n \}$ I claim $D$ is dense in $X$.
Let $x\in X$ and $U$ an open set containing $x$. Hence there exists a basis member, $B$, such that $x\in B=\bigcap_{j=1}^m\pi_{n_j}^{-1}(U_{n_j})\subseteq U$
Hence, $x_{n_p}\in U_{n_p}$ for $p=1,2...,m$. So, $U_{n_p}\cap D_{n_p}\neq \varnothing$.
Then, pick $y_{n_p}\in U_{n_p}\cap D_{n_p}$ for each p=1,2...,m
So, $\textbf{y}\in B\subseteq U$. Therefore, $U\cap D\neq \varnothing$.
Is this correct?
The obvious idea is to try $D=\prod_nD_n$, but of course this fails: it’s certainly dense in $X$, but it’s also uncountable if infinitely many of the sets $D_n$ contain at least two points. The trick is to include enough of this product to make the result dense in $X$ without including so much that the dense set is uncountable. In particular, we need to ensure that it meets every basic open set in the product.
Each of those basic open sets is determined by what happens on a finite set of coordinates: if $F\subseteq\Bbb N$ is finite, and $U_n$ is an open set in $X_n$ for each $n\in F$, whether a point $x=\langle x_n:n\in\Bbb N\rangle$ is in $\bigcap_{n\in F}\pi_n^{-1}[U_n]$ depends on the finitely many coordinates $x_n$ for $n\in F$. If we make sure that $D$ contains a point $y=\langle y_n:n\in\Bbb N\rangle$ such that $y_n\in\pi_n^{-1}[U_n]$ for each $n\in F$, we’ll have ensured that the basic open set $\bigcap_{n\in F}\pi_n^{-1}[U_n]$ intersects $D$.
The most straightforward way to do this is to make sure that for each finite $F\subseteq\Bbb N$ and each $\langle d_n:n\in F\rangle$ such that $d_n\in D_n$ for each $n\in F$ there is a point $x=\langle x_n:n\in\Bbb N\rangle\in D$ such that $x_n=d_n$ for each $n\in F$. The other coordinate of $x$ don’t matter, but we do have to specify them. An easy way to do that is to begin by fixing a point $p=\langle p_n:n\in\Bbb N\rangle$ in $X$ arbitrarily and using its coordinates when we need to fill in a coordinate that doesn’t matter. We can write this up as follows.
Fix $p=\langle p_n:n\in\Bbb N\rangle\in X$ arbitrarily. For each non-empty finite $F\subseteq\Bbb N$ and each $\bar d=\langle d_n:n\in F\rangle\in\prod_{n\in F}D_n$ let $x_{F,\bar d}=\langle x_{F,\bar d,n}:n\in\Bbb N\rangle\in X$, where
$$x_{F,\bar d,n}=\begin{cases} d_n,\text{if }n\in F\\ p_n,\text{otherwise.} \end{cases}$$
Let
$$D_F=\left\{x_{F,\bar d}:\bar d\in\prod_{n\in F}D_n\right\}\;,$$
and let $D=\bigcup\{D_F:F\text{ is a non-empty finite subset of }\Bbb N\}$.
To see that $D$ is dense in $X$, let $F$ be any non-empty finite subset of $\Bbb N$, and for each $n\in F$ let $U_n$ be a non-empty open set in $X_n$, so that $\bigcap_{n\in F}\pi_n^{-1}[U_n]$ is an arbitrary basic open set in $X$. For each $n\in F$ there is a point $d_n\in D_n\cap U_n$; if we take $\bar d=\langle d_n:n\in F\rangle$, then $x_{F,\bar d}\in D\cap\bigcap_{n\in F}\pi_n^{-1}[U_n]$. Thus, $D$ meets every basic open set in $X$ and is therefore dense in $X$.
Finally, we must show that $D$ is countable. If $F\subseteq\Bbb N$ is finite, then $\prod_{n\in F}D_n$ is the Cartesian product of finitely many countable sets and is therefore countable, so $D_F$ is countable. And $\Bbb N$ has only finitely many finite subsets, so $D$ is the union of countably many countable sets and is therefore also countable.
It is possible to avoid having to use the fact that $\Bbb N$ has only countably many finite subsets. Instead of using all non-empty finite subsets of $\Bbb N$, we can get away with using just the ones of the form $F_n=\{0,1,\ldots,n\}$ and taking $\hat D=\bigcup_{n\in\Bbb N}D_{F_n}$, which is clearly the union of countably many countable sets. To see that $\hat D$ is still dense in $X$, let $F$ be any non-empty finite subset of $\Bbb N$, and let $U_n$ be a non-empty open set in $X_n$ for each $n\in F$. Let $m=\max F$ and pick $\bar d=\langle d_n:n\le m\rangle\in F_m$ as follows: if $n\in F$, choose $d_n\in D_n\cap U_n$, and if $n\in F_m\setminus F$, let $d_n$ be any point of $D_n$. It is now easy to check that
$$x_{F_m,\bar d}\in D_{F_m}\cap\bigcap_{n\in F}\pi_n^{-1}[U_n]\subset\hat D\cap\bigcap_{n\in F}\pi_n^{-1}[U_n]\;,$$
as desired.