Show that if $X$ is a countable product of spaces having countable dense subsets, then $X$ has a countable dense subset.
My attempt: Fix $p=(p_n)_{n\in \Bbb{N}} \in \prod_{n\in \Bbb{N}} X_n$. Let $D_n$ be countable dense subset of $X_n$, $\forall n\in \Bbb{N}$. Define $E_n =\{ (x_n)_{n\in \Bbb{N}}| x_i \in D_i, \forall i\in J_n \wedge x_i=p_i, \forall i\gt n\} =(\prod_{i=1}^{n}D_i) \times (\prod_{i=n+1}^{\infty}\{p_i\})$. Claim: $D=\bigcup_{n\in \Bbb{N}} E_n$ is countable dense subset of $\prod_{n\in \Bbb{N}}X_n$. Proof: $\exists$ a bijection between $E_n$ and $\prod_{i\in J_n} D_i$. Map $f: \prod_{i\in J_n} D_i \to E_n$ defined by $f(\prod_{i\in J_n} d_i)= \prod_{i\in \Bbb{N}}e_i$, where $e_i=d_i$, $\forall i\in J_n$ and $e_i=p_i$, $\forall i\gt n$. Map $g: E_n \to \prod_{i\in J_n} D_i$ defined by $g(\prod_{i\in \Bbb{N}} e_i)=\prod_{i\in J_n}e_i$. It’s easy to check $f\circ g=\text{id}_{E_n}$ and $g\circ f=\text{id}_{\prod_{i\in J_n} D_i}$. Thus $f: \prod_{i\in J_n} D_i \to E_n$ is bijective/invertible. By theorem 2.13 of Baby Rudin, $\prod_{i\in J_n} D_i$ is countable. So $E_n$ is countable. By theorem 2.12 of Baby Rudin, $D=\bigcup_{n\in \Bbb{N}} E_n$ is countable.
Let $x=(x_i)_{i\in \Bbb{N}}\in \prod_{i\in \Bbb{N}} X_i$ and $V\in \mathcal{N}_x$. Since $x\in V\in \mathcal{T}_p$, we have $\exists B\in \mathcal{B}_p$ such that $x\in B \subseteq V$. By definition of basis, $B=\prod_{i\in \Bbb{N}}U_i$, where $U_i =X_i$, $\forall i\in \Bbb{N}\setminus F$ and $U_i \in\mathcal{T}_{X_i}$, $\forall i\in \Bbb{N}$. Let $m= \max F$. $x_i \in U_i$, $\forall i \in J_m$. So $U_i \in \mathcal{N}_{x_i}$, $\forall i\in J_m$. Since $x_i \in X_i =\overline{D_i}$, we have $D_i \cap U_i \neq \emptyset$. So $\exists z_i \in D_i \cap U_i$, $\forall i\in J_m$. Construct $q=(q_i)_{i\in \Bbb{N}}$ such that $q_i=z_i$, $\forall i\in J_m$ and $q_i=p_i$, $\forall i\gt m$. Then $z\in E_m$ and $z\in B\subseteq V$. So $z\in E_m \cap V \neq \emptyset$. Since $E_m \cap V \subseteq (\bigcup_{n\in \Bbb{N}}E_n) \cap V= D\cap V$, we have $D\cap V\neq \emptyset$. Thus $x\in \overline{D}$. Hence $\overline{D}=\prod_{i\in \Bbb{N}}X_i$. Is this proof correct?
Is there any alternative construction of $D$?