The extended real line can be given the order topology i.e the topology generated by the basis elements $(a,b)=\{a<x<b\},R_a=\{x>a\}$ and $L_a=\{x<a\}$ where $a,b\in \bar R$.
Hence, a countable basis would just be the above sets except with the restriction that $a,b\in \mathbb Q\cup \{+\infty,-\infty\}$
Best Answer
Yes, that's true. Restricting the base elements to rational endpoints (in all 3 types) gives a countable base, as Q is order dense in the extended reals: for any $a < b$ we can find a rational strictly in-between.
As all order topologies are $T_3$ (even $T_5$) Urysohn's metrication theorem already tells us it's metrisable (which is also obvious if you know that $\bar{\Bbb R}$ is homeomorphic to $[-1,1] \subseteq \Bbb R$.)