Proof Attempt: Let ($X$, $\mathscr{T}$) be a topological space and $\mathscr{B}$ be the base of $\mathscr{T}$ such that $\mathscr{B}=\{B_i\}_{i\in\Bbb{N}}$. We want to show that $\exists A\subset X$ such that $X$ is the closure of $A$ and $A$ is countable.
Let $x\in X\setminus A$. If $\overline{A}=X$, then for every neighborhood $\mathscr{O}_x$ of $x$, $\exists \alpha \in A\setminus\{x\}$ such that $\alpha \in \mathscr{O}_x$. Note $A\setminus\{x\}=A$ and $\exists k\in\Bbb{N}$ such that $x \in B_k \subset \mathscr{O}_x$. Since $B_k$ is also a neightborhood of $x$, then $\exists \alpha_k \in A\setminus \{x\}$ s.t. $\alpha_k \in B_k$. Thus, $B_k \cap A\setminus \{x\}\neq \emptyset$. So for any $x' \in X\setminus A$ and for each $B_i\in\mathscr{B}$ that contains $x' \in X\setminus A$, pick one $a'\in A\setminus \{x'\}=A$ such that $a' \in B_i$ and collect them in a set $A'$; that is, $A'$ is the set of all those chosen $a'$'s. Keep in mind that $\vert A'\vert\leq\Bbb{N}$ so $A'$ is countable.
Furthermore, we consider the accumulation points $A_{cc}$ of $A$ within $A$. Note that either $A\setminus A_{cc}$ is countable or uncountable. Suppose $A$ is uncountable. By Theorem 9, if $A\setminus A_{cc}$ is still uncountable, then there there exists accumulation points of $A\setminus A_{cc}$ within itself. Since $A\setminus A_{cc}\subset A$, that mean there still exists accumulation points of $A$ in $A\setminus A_{cc}$ which is a contradiction as $A_{cc}$ is not in $A\setminus A_{cc}$. This forces $A\setminus A_{cc}$ to be countable. Observe that $\overline{A \setminus A_{cc}}=A$. Since the union of countable sets is countable, $(A \setminus A_{cc})\cup A'$ is countable. By our constructed definition of $A\setminus A_{cc}$ and $A'$, we have $\overline{(A \setminus A_{cc})\cup A'}=X$.
Finally, by picking a point for each basis element, the closure of such set of points is $X$ and as shown earlier this set of points must be countable. Hence, $X$ is separable.
For each base set $B_n$, pick some $a_n$ from $B_n$.
Let $A= \{ a_n : n \in \mathbb{N} \}$. Clearly $A$ is countable.
It is not difficult to show $A$ is dense, its closure $X$.