Let $D$ be a dense subset of the metric space $X$ with metric $d.$ Let $B=\{B_d(p,q) :p\in D \land q\in Q^+\}.$ Then $B$ is a base for $X.$ To prove this it suffices to show that whenever $y\in U$ with $U$ open, there exists $ V\in B$ with $ y\in V\subset U.$ Consider that $B_d(y,r)\subset U$ for some $r>0$, and that some $p\in D\cap B_d(y,r/3).$ Take $q\in Q^+$ with $r/3<q<r/2.$ By the triangle inequality, $y\in B_d(p,q)\subset B_d(y,r)\subset U.$...... In particular if $D$ is countable then so is $B$. The set of members of $R^n$ with rational co-ordinates is countable, and dense in $R^n$.
Let $(X,d)$ be a metric space and let $E$ be a dense subset of $X.$ Let $B=\{B_d(x,q): x\in E\land q\in \mathbb Q^+\}.$ Then $B$ is a base for $X.$
Proof: Let $p\in U$ where $U$ is open in $X.$ It suffices to show that $p\in b\subset U$ for some $b\in B.$
There exists $r>0$ such that $B_d(p,r)\subset U,$ and there exists $q\in (0,r/2)\cap \mathbb Q^+.$ And there exists $e\in E$ with $d(p,e)<q/2$ because $E$ is dense in $X.$ So $p\in B_d(e,q)\in B.$
And $y\in B_d(e,q)\implies d(y,p)$ $\leq d(y,e)+d(e,p)<q+q/2<2q<r$ $\implies y\in B_d(p,r)\subset U.$ So $p\in B_d(e,q)\subset U.$.... Q.E.D.
If $E$ is countable then $B$ is countable. When $X=\mathbb R^n$ with the usual topology, let $E$ be the set of points with rational co-ordinates. Then $E$ is dense and countable so $B$ is a countable base for $\mathbb R^n.$
EDIT: formatting second paragraph in proof
Best Answer
Let $(u_n)$ be a descending sequence of rationals converging to $a$.
Let $(v_n)$ be an ascending sequence of rationals converging to $b$.
$$(a,b) = \bigcup_n (u_n, v_n)\text{.}$$