Why there is $\varepsilon>0$ such that $\sum_{i=1}^\infty |Q_i|\leq m^*(E)+\varepsilon$. Why for all $E\subset \mathbb R^n$ there is a cover $E\subset \bigcup_{i=1}^\infty Q_i$ of closed cube such that $$\sum_{i=1}^\infty |Q_i|\leq m^*(E)+\varepsilon$$
I know that $$m^*(E)=\inf\left\{\sum_{i=1}^\infty |Q_i|\mid E\subset \bigcup_{i=1}^\infty Q_i\right\}$$
where $Q_i$ are closed cube. So I by the property of the infimum it's clear that if $\varepsilon>0$ there is $\bigcup_{i=1}^\infty Q_i\supset E$ such that $$m^*(E)\leq \sum_{i=1}^\infty |Q_i|\leq m^*(E)+\varepsilon,$$
but the question is why such a cover $\bigcup_{i=1}^\infty Q_i\supset E$ exist ? If my question is not clear, tell me.
 A: I'm under the impression that you are mixing up things a bit.
i) You almost certainly don't want to find an $\varepsilon > 0$ with this property, but you want, for any given $\varepsilon > 0$ find a family of cubes such that your statement becomes true.
ii) the $\inf$ in the definition of $m^*(e)$ is the infimum over all families of cubes covering $E$. So in fact, by the definition, you will find such a cover. 
Edit: if the value of $m^*(e) $ is $ \infty $ note that $ \infty + $ any finite value is just $ \infty $ .
(2nd Edit: the question was modified after I wrote this answer, so i) does not make sense any more, but I won't write the answer from scratch.... )
A: Let $E\subset \mathbb R^n$ a bounded set and $\varepsilon>0$. The set $$\mathcal O=\left\{x\in\mathbb R^n\mid d(x,E)<\varepsilon\right\}$$
is an open set that contain $E$. Moreover, all open set of $\mathbb R^n$ can written as a countable union of almost disjoint closed cube, i.e. $$\mathcal O=\bigcup_{i=1}^\infty Q_i$$ where $Q_i$ are closed cube almost disjoint (i.e. $\text{int}(Q_i)\cap\text{int}(Q_j)=\emptyset$ for all $i\neq j$). 
Therefore, you can always approche $E$ by a countable union of closed cube.
