Measure and Integration Theory

Reason directly: given a countable sequence of intervals covering $(0,1)$, find a finite subcover and proceed as in the proof that $m^\ast([a,b]) \geq b-a$.

Excision: $m^\ast((0,1)) \geq m^\ast([0,1]) - m^\ast(\{0\}) - m^\ast(\{1\})$ by subadditivity of $m^\ast$.

Approximate by closed sub-intervals: $m^\ast((0,1)) \geq m^\ast([\epsilon,1-\epsilon])$ for all sufficiently small $\epsilon>0$.