Set of open intervals in R with rational endpoints is a basis for standard topology on R 
Show that the set $\mathcal{B} = \{(a,b) \subset \mathbb{R}: a,b \in \mathbb{Q}\}$ is a basis for the standard topology on $\mathbb{R}$

First I'll show that $\mathcal{B}$ is a basis on $\mathbb{R}$ and then i'll do a fineness comparison via bases:

1) Since $\cup_{p,q \in \mathbb{Q}\;for \;p < q}(p,q) = \mathbb{R}$, $\mathcal{B}$ covers $\mathbb{R}$. Since the intersection of open intervals is an open interval, every point in the intersection of two open intervals is contained in an open subinterval. Thus, $\mathcal{B}$ is a basis for some toplology $\tau$ on $\mathbb{R}$.2) Let$\;\mathcal{B_{stand}}\;$be the standard basis for the standard topology $\tau_{stand}$ on $\mathbb{R}$. Let $(p, q)\in \mathcal{B}$. Since $\cup_{x \in \mathbb{R},\;x > p} (x,q) = (p,q)$, every basis element in $\mathcal{B}$ can be written as the union of basis elements in $\mathcal{B_{stand}}$. Thus, $\tau_{stand}$ is finer than $\tau$.Let $(x, y)\in \mathcal{B}$. Since $\cup_{p,q \in \mathbb{Q}, \;x< p< q < y\;} (p,q) = (x,y)$, every basis element in $\mathcal{B_{stand}}$ can be written as the union of basis elements in $\mathcal{B}$. Thus, $\tau$ is finer than $\tau_{stand}$.$\therefore\;$ $\mathcal{B} = \{(a,b) \subset \mathbb{R}: a,b \in \mathbb{Q}\}$ is a basis for the standard topology on $\mathbb{R}$

Feedback on my proof and alternative ways to solve the problem would be appreciated.
(I also found a post where someone asked the same question, but I'm looking for  proof verification/alternative proofs so I'm not sure what the protocol is.)
Collection of open intervals $(a,b) \ a,b\in \mathbb{Q}$ is a basis for euclidean topology on $\mathbb{R}$
 A: A couple of thoughts. First of all when you say "Since the intersection of open intervals is an open interval, every point in the intersection of two open intervals is contained in an open subinterval. Thus, $\mathcal{B}$ is a basis for some topolology $\mathcal{T}$ on $\mathbb{R}$." you need to be a bit more specific. You want to show specifically that if you have 2 basis elements $B_1$ and $B_2$ that both contain a point $x$ then you can pick a third basis element that contains $x$ and is in $B_1 \cap B_2$. To translate back to this problem you want to show that your open interval you get from intersecting the other 2 (or some smaller one inside it) has rational endpoints.
Using your notation for comparing bases we can show that $\mathcal{B}_{stand}$ is finer than $\mathcal{B}$ just by showing that $\mathcal{B} \subset \mathcal{B}_{stand}$. This is true because open intervals with rational endpoints are also just open intervals.
Finally I personally think that this lemma from Munkres would make for a more elegant argument than the final part of yours. 
Lemma: $\mathcal{T}'$ is finer than $\mathcal{T}$ if and only if for each $x \in X$ and every basis element $B \in \mathcal{B}$ there is a basis element $B' \in \mathcal{B}'$ such that $x \in B' \subseteq B$.
Then instead of constructing any open interval as the union of open intervals with rational endpoints you just need to find one open interval with rational endpoints in this open interval that contains this arbitrary point $x$ from the interval.
