I'm trying to understand the validity of this claim and prove it:
if $E \subset \mathbb R^n$ is compact then $\forall \epsilon >0$ there is a finite covering $E \subset \cup_{j=1}^{m} Q_j$ such that $\sum_{j=1}^{m}v(Q_j) < \epsilon$
But wouldn't make $E$ a null-set? isn't that the definition of a set that has zero measure?
But for instance, $[0,1] \subset \mathbb R$ is compact, but it is not zero measure. Is this statement even true?
As pointed out in the comments, your statement is wrong. If $E$ is compact, the only thing you can say is that for any $\epsilon >0$ you can cover $E$ with a finite family of sets such that each of them has measure less than $\epsilon$.
Observe that the cardinality of the covering actually depends on $\epsilon$. Therefore you can't conclude that the measure of $E$ is zero.