There are questions on Math.SE that seem relevant, but I will explain why they do not answer my question:
1) $\mathbb R^n$ has countable basis of open balls? (Yes): this question doesn't prove that rational centre/radii balls form a base; it proves that $\mathbb{R}^n$ has such a base
2) Open ball with rational radii forms a basis.: makes a claim that I do not know how to show (it is not obvious to me):
Now let $B$ be an open ball with centre $a$, and irrational radius $\rho$. Then $B$ is the union of all the open balls with centre $a$ and rational radius $r\lt \rho$.
3) How to cover an open subset of $\mathbb{R}^n$ with balls?: the answer to this question gets me the closest, I think, but I am still having trouble with it. Let me break it down my issues:
HINT: Every open set in $\mathbb{R}^n$ is a union of open (or closed) balls whose centres have rational coordinates and whose radii are rational; how many such balls are there?
Alright, assuming that I can show that, I still don't know how to show its countable. My guess is I have to play around and find a bijective function that connects each ball with an integer. Is this the right approach?
In case the first statement isn’t obvious, suppose that $B$ is the ball of radius $r$ about a point $x\in\mathbb{R}^n$. If $x$ has all rational coordinates, there’s nothing to be done.
How is there nothing to be done? Don't we still have to show that $r$ (the radius) is rational?
Otherwise there is a point $y$ with all coordinates rational inside the ball of radius $r/2$ centred at $x$.
Why is this an obvious statement? I don't know how to begin to show it. Could a hint be provided here?
Let $d$ be the distance between $x$ and $y$, let $q$ be a rational number such that $d<q<r/2$, and let $B'$ be the ball of radius $q$ centred at $y$; then $x\in B'\subseteq B$.
This part I am totally good with.
Anyway, so if I can show that there a countable number of such balls, then by definition of second-countability, $\mathbb{R}^n$ is second countable.
Best Answer
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