We want to see that $\mathscr{B}$ has the cardinality of $\mathbb{R}$, that is to say, the cardinality of continuum $c$.
Here: Cardinality of Borel sigma algebra
, there is a proof of the fact that if a sigma algebra is generated by countably many sets, then the cardinality of it is either finite or $c$.
It is clear that it is the same than $\mathbb{R}$ or $\mathscr{P}(\mathbb{R})$, so it isn´t finite.
By definition $\mathscr{B}$ is the sigma algebra generated by the usual topology:
$\mathscr{B}=\sigma(\tau_{us})$. Every open set $A$ in $\mathbb{R}$ can be seen as the union of the open (basic) neighbourhoods of the points in $A \cap \mathbb{Q}$ since $\mathbb{Q}$ is dense , that is,a numerable union of sets in $\tau_{us}$. So the collection
$$ \mathscr{A}=\{(q,p) \quad : q<p \quad ; \, q,p \in \mathbb{Q}\}$$
is countable and satisfy : $ \tau_{us} \subset \sigma(\mathscr{A}) $, so:
$$\mathscr{B}=\sigma(\tau_{us})=\sigma(\mathscr{A})$$
However, usually $\mathscr{B}$ is presented as $\sigma(\{(-\infty,a] : a\in \mathbb{R}\})$
And it is easy to see that
$$\{(-\infty,a] : a\in \mathbb{R}\} \quad \subset \quad \sigma(\{(-\infty,q] : q\in \mathbb{Q}\})$$
since $(-\infty,a]=\bigcap_{q\geq a ,q\in \mathbb{Q}}(-\infty,q]$ ,
so the sigma-algebra which generates the first ($\mathscr{B}$) is the same that the one which generates the second (countable) set.
I don´t know what you mean by "a way to see this through the concept of standard topology". Actually, the fact that this collection generates $\mathscr{B}$ rely on the nature of the topology of $\mathbb{R}$ : it is second countable.
You can see many of the properties of Borel Algebras in these topology spaces in several posts, for example:
cardinality of the Borel $\sigma$-algebra of a second countable space