representing $E$ as the disjoint union of a finite number of measurable sets that have a measure of at most $\epsilon > 0$

Solution 1:

That proof looks fine, but it is a bit long. Consider: If $x$ arbitrary then consider $B_\rho(x)\cap E$. Suppose $m(E)>\epsilon$. Then there exists a $\rho$ so that $m(B_\rho(x)\cap E)=\epsilon$. Thus we can write $E$ as disjoint union $$ E= (B_\rho(x)\cap E)\cup (E\setminus (B_\rho(x)\cap E)) = F_1\cup E_1$$ where one set has measure $\epsilon$ and the other $m(E)-\epsilon$. Thus by iterating this argument $N=[m(E)/\epsilon]$ times we get $m(E_N)<\epsilon$, by which we get such a partition.

Disjoint union means that all sets are pairwise disjoint. This is not really important here, as if we have a union $E=\bigcup F_i$ we can replace $F_i$ by $F_i\setminus \bigcup_{j<i} F_j$ to obtain a disjoint union where each set has at most the measure of the original set (since $F_i\setminus \bigcup_{j<i} F_j\subseteq F_i$).