I believe this follows from the Urysohn metrization theorem. Order topologies are regular (even completely normal, according to Wikipedia), and your hypothesis of a dense countable subset should imply second countability, unless I'm missing something obvious. Namely open intervals with endpoints in your dense set will form a basis for the topology. So the two hypotheses of Urysohn's metrization theorem are satisfied.

I believe this follows from the Urysohn metrization theorem. Order topologies are regular (even completely normal, according to Wikipedia), and your hypothesis of a dense countable subset should imply second countability, unless I'm missing something obvious. Namely open intervals with endpoints in your dense set will form a basis for the topology. So the two hypotheses of Urysohn's metrization theorem are satisfied.

Edit: This works if you have a countable dense subset $D$ in the sense that for every $x$ and $y$ there exists a $d\in D$ with $x < d < y$, but that's not the sense of density being used!