How do I show $P(\mathbb Q) = \mathscr B(\mathbb R) \cap \mathbb Q$ My first question is to confirm that $\sigma$-algebra generated by the set $\{ (a,b] \cap \mathbb Q \mid -\infty \leq a \lt b \leq \infty \}=\mathscr B(\mathbb R) \cap \mathbb Q$
where $\mathscr B$ is the Borel $\sigma$-algebra.
Than how do I show $P(\mathbb Q) = \mathscr B(\mathbb R) \cap \mathbb Q$,
I think it is pretty clear that $\mathscr B(\mathbb R) \cap \mathbb Q \subset P(\mathbb Q)$ as $\mathbb Q \subset P(\mathbb Q)$. How about the other direction?
 A: Hint:
Given $S \subset \Bbb Q$, what happens if you take the union of singletons $\{x\}$ for all $x\in S$? 
Each singleton is closed, and $\mathscr B(\mathbb R)$, the Borel $\sigma$-algebra, has all open and closed sets.

The comment is getting a bit long:
Only finite unions of closed sets are guaranteed to be closed, so you can't conclude $S$ is closed and it's not necessarily: take a sequence of rationals converging to an irrational, for example. It has a limit point, but it does not belong to the set of elements of the sequence.
But you can however conclude it's a Borel set, because $\sigma$-algebras are stable by countable unions, and $S$ must be countable since it's a subset of $\Bbb Q$.
I use $S$ because I need to prove that for all $S\subset \Bbb Q$, $S$ is also in $\mathscr B(\mathbb R)$, thus $P(\Bbb Q) \subset \mathscr B(\mathbb R)$.

By the way, using the same argument, you can prove that any countable subset of $\Bbb R$ is in $\mathscr B(\mathbb R)$.
However, there are (uncountable) subsets of $\Bbb R$ which are not Borel sets, see wikipedia or here on MathOverflow. There are probably also questions about this here on MSE.
