Let $\tau$ be the usual topology of $\mathbb{Q}$. As $\tau$ is a collection of subsets of $\mathbb{Q}$, we can see $\tau$ as a subset of $2^{\mathbb{Q}}$, which is a Polish space.
I want to show that $\tau$ is a Borel set of $2^{\mathbb{Q}}$. Exactly, I want to show that $\tau$ is $\Pi^0_3(2^{\mathbb{Q}})$.
I think the following characterization of $\mathbb{Q}$ can be useful.
It's well-known that $\mathbb{Q}$ is a unique, up to homeomorphism, countable perfect metric space.
Let $B_n(q)$ be the ball of radius $\frac{1}{n}$ centered at $q$. Then
$$ X \in \tau \iff \forall q \in \mathbb{Q} . \exists n \in \mathbb{N} . \forall r \in B_n(q) . [ q \in X \implies r \in X] $$
Put differently,
$$ \tau = \bigcap \limits_{q \in \mathbb{Q}} \bigcup \limits_{n \in \mathbb{N}} \bigcap \limits_{r \in B_n(q)} N_{q,r} $$
Where $N_{q,r} = \{ X ~|~ q \in X \implies r \in X \} = \{ X ~|~ q \not \in X \lor r \in X \}$ is clopen.
Since $\mathbb{Q}$, $\mathbb{N}$, and $B_n(q)$ are all countable, $\tau$ is $\Pi^0_3$, as desired.
Hope this helps ^_^