Yes, in fact every countable ordinal embeds into the rational numbers in this way, an order-preserving map as a closed set of rational numbers.

One can easily prove this by induction on ordinals.

Alternatively, one can also prove it using the universal property of the rational order. For any countable ordinal $\alpha$, extend the order of $\alpha$ to a dense linear order by adding a copy of $\mathbb{Q}$ between every ordinal and its successor. This gives a countable dense linear order, which must be order isomorphic to an interval of $\mathbb{Q}$. The original order is closed in the larger order, so this gives the ordinal as a closed subset of $\mathbb{Q}$.

Yes, in fact every countable ordinal embeds into the rational numbers in this way, an order-preserving map as a closed set of rational numbers.

Let me give three proofs.

First, one can easily prove this by induction on ordinals. Having a closed copy of $\alpha$ in $\mathbb{Q}$, we can scale it down to a bounded interval and then add another point on top to get a copy of $\alpha+1$. And if we have copies of every $\alpha<\lambda$ for a countable limit ordinal, we can similarly scale to successive intervals in such a way that the combined embedding contains its limit points.

Second, alternatively, one can also prove it using the universal property of the rational order. For any countable ordinal $\alpha$, extend the order of $\alpha$ to a dense linear order by adding a copy of $\mathbb{Q}$ between every ordinal and its successor. This gives a countable dense linear order, which must be order isomorphic to an interval of $\mathbb{Q}$. The original order is closed in the larger order, so this gives the ordinal as a closed subset of $\mathbb{Q}$.

Third, here is another soft proof. Cantor proved that the rational order $\mathbb{Q}$ is universal for all countable linear orders. So every countable ordinal $\alpha$ admits an order-embedding into $\mathbb{Q}$. The image of this embedding might not be closed in the way you have asked. But let us simply add the limit points. The new set (the old copy of $\alpha$ plus the limits) will still be well ordered and thus be a copy of some ordinal $\beta\geq \alpha$. So taking the first $\alpha$ many elements of this set gives a closed copy of $\alpha$ in $\mathbb{Q}$.