Let $(X,T)$ be a topological space. The claim is that $E\in\mathscr{B}(X):=\sigma(T)$ if and only if $E$ can be obtained by countable union/intersection/complement of open sets in $X$.
For the reverse direction, if $E$ is a countable union/intersection/complement of open sets in $X$, then $E\in S_\alpha$ for each $\sigma$-algebra $S_\alpha$ such that $T\subseteq S_\alpha$, so $E\in\mathscr{B}(X)$.
But I'm not sure how to prove the other direction. If $E\in\mathscr{B}(X)$, then $E\in S_\alpha$ for every $\sigma$-algebra $S_\alpha$ such that $T\subseteq S_\alpha$. Why should $E$ be a countable union/intersection/complement of open sets?
The set of all rational numbers is not a countable union/intersection of complement of open sets. The fact that it is not a countable intersection of open sets is not elementary, but it is a standard application of Baire Category Theorem. Of course this set is a Borel set because it is a countable union of singleton sets.