 Solution

Every open interval is open in the usual metric topology. Conversely, if $U\subset \mathbb{R}$ is open and $x\in U$, there exists ${ϵ}_x>0$ with $(x-{ϵ}_x,x+{ϵ}_x)\subseteq U$. Taking the union of these intervals over all $x\in U$ recovers $U$. Thus open intervals form a basis of the usual topology.