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 $\mathbb{Q}$.