Let $\mathscr{B}=\{[a,b):a\in\mathbb{R}, b\in\mathbb{Q},a<b\}$.
Prove that $\mathscr{B}$ is a basis for the topology $\tau_1$ in $\mathbb{R}$.
I am having a hard time trying to prove this claim. I can intuitively understand that $\mathscr{B}$ is a basis. However it has been difficult to prove.
I need to prove the union of elelments of the basis equals $\mathbb{R}$ but I have no clue on how to do it.
Regarding the intersection being the union of elements of the basis I attempted the following:
If $c,d\in\mathbb{R}$ and $e,f\in\mathbb{Q}$ such that $c<d<e<f$. Then $[c,e),[d,f)\in\mathscr{B}$, such that $[c,e)\cap[d,f)=[d,e)$. However $\exists h\in [d,e)$ such that $[d,h)\cup [h,e)=[d,e)$.
Question:
1) How should I prove $\mathscr{B}$ to be a basis of the Sorgenfrey line?
2) On my attempted proof I have a problem regarding $h$. Once I cannot assume it belongs to $\mathbb{Q}$.
Thanks in advance!
If $O$ is Sorgenfrey open and $x \in O$, we know that for some $y \in \mathbb{R}$ we have that $[x,y) \subseteq O$ (as the half-open half-closed intervals form the standard base or generating set for the Sorgenfrey topology). Now by order density of the rationals (which follows from the construction/definition of the real numbers in the first place) there is some $q \in \mathbb{Q}$ with $x < q < y$.
But then $[x,q) \in \mathcal{B}$ and $[x,q) \subseteq [x,y) \subseteq O$.
So we have shown that for all open $O$ and all $x \in O$, we have some $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq O$. This by definition means that $\mathcal{B}$ is a base for the Sorgenfrey line.
It implies easily e.g. that $O = \bigcup \{B_x: x \in O\}$ (as every $x \in O$ is in its "own" $B_x$ and conversely all $B_x \subseteq O$ hence so is their union, so all open $O$ are unions of base elements).