Proof of subadditivity of the Lebesgue measure

$m^*(E_n)=\inf{\left\{\displaystyle\sum_{k=0}^\infty \ell(I_k)\mid E_n\subseteq\bigcup_{i=0}^\infty I_k \right\}}$, where $I_k$ is an open bounded interval. Since what is on the left hand side of your inequality is one of these possible coverings, the infimum, $m^*(E_n)$ is less than or equal to that. This means you can find an $\epsilon>0$ such that $m^*(E_n) + \epsilon > \displaystyle\sum_{k=0}^\infty \ell(I_k)$, where $\{I_k\}_{k=0}^\infty$ is one possible colleciton of bounded open sets containing $E_n$. Since we have a countable collection of these sets, $E_n$, we can associate to each set a natural number $n$ so that your inequality holds for the choice of $\dfrac{\epsilon}{2^n}$ for the $n$th set.