Metric Topology Suppose our topological space is $\mathbb{R}$. Why is each basis element $(a,b)$ for the order topology is a basis element for the metric topology? Munkres says, $(x-\epsilon,x+\epsilon)=B(x,\epsilon)$ 
Is this because by definition,
$B(x,\epsilon)=\{y:d(x,y)<\epsilon\}$ and $(x-\epsilon, x+\epsilon)=\{y:|y-x|<\epsilon\}$
And in our usual topology on $\mathbb{R}: d(x,y)=|x-y|.$

The reason you give that $B(x,\varepsilon) = (x - \varepsilon, x+\varepsilon)$ are essentially correct, yes. This shows that every metric ball is an open interval (and thus open in the order topology). If $(a,b), a < b$ is any open interval, setting $m = \frac{a+b}{2}$ (the midpoint) and $r = \frac{b-a}{2}$ (the distance between the endpoints halved), it's easy to see that $(a,b) = (m - r, m+ r) = B(m,r)$, so every open interval is open in the metric topology as well. This means that the topologies are equal: the base elements of one are open in the other and vice versa, so every order open set (a union of open intervals ) is metric open and vice versa.
