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 $\mathbb{Q}$.