I am aware that this question has been asked multiple times on this site; however none of them seem to have a proof as simple as mine. So I would like to know if this is legitimate:
Let $A=\{x_n:n\in\mathbb Z_+\}$ be a countable dense set in $X$, then $\overline A=X$. So if $x\in X$, then $x\in A$ or $x\in \overline A$. Either way, any neighborhood $U$ of $x$ contains a point of $A$. So there exist positive integers $n,m$ such that $B_d(x_n,1/m)\subset U$, with $x\in B_d(x_n,1/m)$. Hence the set $$B =\{B_d(x_n,1/m):n,m\in\mathbb Z_+\}$$
is a countable basis for $X$.
Best Answer
Let $d$ be the metric on $X$. Let $A=\{x_n:n\in\mathbb Z_+\}$ be a countable dense set in $X$, then $\overline A=X$. Put $$\mathcal B = \{B_d(x,r):x\in A,r\in\mathbb Q\},$$ then $\mathcal B$ is countable as $A$ and $\mathbb Q$ are countable. Let $U$ be an open set in $X$ and $x\in U$. Because $U$ is open, we may choose $\varepsilon>0$ such that $B_d(x,\varepsilon)\subset U$, and since $A$ is dense there exists $x_n\in A$ with $x_n\in B_d(x,\varepsilon/3)$. Pick $r\in\mathbb Q$ with $\varepsilon/3<r<\varepsilon/2$, then $x\in B_d(x_n,r)$ since $B_d(x_n,r)\subset B_d(x,\varepsilon/3)$. Moreover, $B_d(x_n,r)\subset B_d(x,\varepsilon)$, since if $d(y,x_n)<r$ then $$ d(x_n,x) \leqslant d(y,x_n) + d(x_n,x) < r + r < \varepsilon. $$ It follows that $x \in B_d(x_n,r)\subset B_d(x,\varepsilon) \subseteq U$, so that $\mathcal B$ is a basis for $X$, as desired, as of course $B_d(x_n,r) \in \mathcal{B}$ by definition.