If it is enough to have all open intervals (a,b) with end points $a$ and $b$ belonging to the rational numbers, a < b, in order to generate a Borel $\sigma$-algebra on $\mathbb{R}$. Asked here: About the open intervals generating a Borel $\sigma$-algebra on $\mathbb{R}$
What kind of numbers do you need to have between $a$ and $b$? Only rational numbers or real numbers? And why?
Following Nate Eldredge, I'll write "$(a,b)_{\mathbb{Q}}$" for $(a,b)\cap\mathbb{Q}$ (with $a<b$ reals; note that we don't need $a,b\in\mathbb{Q}$ themselves for this to make sense). And I'll reserve "$(a,b)$" for the full interval of real numbers, as usual.
Now the key point is that each $(a,b)_\mathbb{Q}$ is countable. This implies that every element $X$ of the $\sigma$-algebra generated by such intervals is either countable or co-countable (= has countable complement). Specifically, let $\mathfrak{S}$ be the set of all sets of reals which are either countable or co-countable; it's easy$^1$ to check that $\mathfrak{S}$ is a $\sigma$-algebra, and it clearly contains each $(a,b)_{\mathbb{Q}}$.
Note $1$: I'm not saying,incidentally, that $\mathfrak{S}$ is the $\sigma$-algebra so generated, merely that it contains it. Indeed, they're not the same, and it's a good exercise to find something in $\mathfrak{S}$ that isn't in the $\sigma$-algebra generated by the $(a,b)_\mathbb{Q}$s.
Note $2$: More generally, it's usually the case that the $\sigma$-algebra generated by a collection of "small" sets consists entirely of "small" or "co-small" sets; e.g. every element of the $\sigma$-algebra generated by the null sets is either null or co-null, every element of the $\sigma$-algebra generated by the meager sets is either meager or co-meager, etc.
But there are plenty of Borel sets which are neither countable nor co-countable - for example, the interval $(0,1)$.
$^1$OK, let's sketch that here.
Complements are immediate: if $A$ is countable (respectively co-countable) then $A^c$ is co-countable (resp. countable).
Countable unions are just complements of countable intersections of complements, so the previous line shows that if $\mathfrak{S}$ is closed under countable intersections it's also closed under countable unions.
So suppose $A_i\in\mathfrak{S}$ for $i\in\mathbb{N}$. We want to show that $X=\bigcap_{i\in\mathbb{N}}A_i$ is either countable or co-countable. There are two cases:
If some $A_i$ is countable, then we can conclude that $X$ is countable; do you see why?
So suppose each $A_i$ is co-countable. The complement of $X$ is the union of the complements of the $A_i$s, which is to say a countable union of countable sets; what does this tell you about $X$?