I have a feeling they are neither closed nor open as the $\mathbb{R} \setminus \mathbb{Q}$ cannot be open or closed either...
In the usual topology of $\mathbb{R}$, $\mathbb{Q}$ is neither open nor closed.

The interior of $\mathbb{Q}$ is empty (any nonempty interval contains irrationals, so no nonempty open set can be contained in $\mathbb{Q}$). Since $\mathbb{Q}$ does not equal its interior, $\mathbb{Q}$ is not open.

The closure of $\mathbb{Q}$ is all of $\mathbb{R}$: every real number is the limit of a sequence of rationals, so every real number lies in the closure of $\mathbb{Q}$. Since $\mathbb{Q}$ does not equal its closure, it is not closed.

Naturally, since $\mathbb{Q}$ is not open, its complement is not closed; since $\mathbb{Q}$ is not closed, its complement is not open.

But this is in the usual topology. $\mathbb{R}$ can be endowed with lots of topologies, and it is certainly possible for $\mathbb{Q}$ to be open (or closed) in some of them. For example, in the discrete topology, where every subset of $\mathbb{R}$ is both open and closed, $\mathbb{Q}$ is both open and closed.
