 Solution

Each interval $(a,b)$ is open in the usual metric: if $x\in (a,b)$, the ball with radius $\min (x-a,b-x)/2$ is contained in it. Conversely, if $U$ is open and $x\in U$, the metric definition supplies $r>0$ with $(x-r,x+r)\subseteq U$. Thus every open set is a union of open intervals, which proves they form a basis.