How does one go about proving the following:
Every open set $A$ in the topological space $(\mathbb{R}^d,\|\cdot\|)$ (with the norm topology) is the union of all the open balls $B_\epsilon(q)$ whose center $q$ and radius $\epsilon>0$ are rational.
In fact, the result that I'm aiming at is to show that $\mathcal{B}(\mathbb{R}^d)$, i.e. the Borel $\sigma-$field in $\mathbb{R}^d,d<\infty$ is the $\sigma-$field generated by the class of sets that are of the form
$$O:=\times_{j=1}^d(a_j,b_j]$$
with $a_j,b_j\in\mathbb{R},\forall j=1,\dots,d$.
Let $U:=\cup\left\{ B_{\varepsilon}\left(q\right)\mid q\in\mathbb{Q}^{d},\varepsilon\in\mathbb{Q}_{>0},B_{\varepsilon}\left(q\right)\subset A\right\} $
Then $U\subset A$.
If $a\in A$ then $B_{\delta}\left(a\right)\subset A$ for some $\delta>0$.
Then $\varepsilon\in\mathbb{Q}_{>0}$ exists with $\varepsilon<\frac{1}{2}\delta$, and $q\in\mathbb{Q}^{d}$ exists with $\left\Vert a-q\right\Vert <\varepsilon$ since $\mathbb{Q}^{d}$ is dense in $\mathbb{R}^{d}$.
Then $a\in B_{\varepsilon}\left(q\right)\subset B_{\delta}\left(a\right)\subset A$.
This because $x\in B_{\varepsilon}\left(q\right)$ implies that: $$\left\Vert x-a\right\Vert \leq\left\Vert x-q\right\Vert +\left\Vert a-q\right\Vert <2\varepsilon<\delta$$
Note that $B_{\varepsilon}\left(q\right)\subset U$ and consequently $a\in U$.
Prove is now that $A=U$.