We say measures $\mu, \nu$ are mutually singular on $E$ if $E$ can be written as disjoint union, $E= E_1\cup E_2$ with $\mu(E_1) = \nu(E_2) = 0$.
Prove that the following condition implies $\nu$ and $\mu$ are mutually singular on $E$: Given any $\epsilon >0$, there are disjoint measurable $E_1$ and $E_2$ with $E=E_1\cup E_2$ and $\mu(E_1)<\epsilon$, $\nu(E_2)<\epsilon$.
I think it might need to use some $\limsup$ of sets, but I don't know how to construct $E_1$ and $E_2$ with measure $0$ respectuvely.
Yes, $\limsup$ is a good idea. Let's use $A$ and $B$ for the sets rather than proliferating indices on $E$.
Choose a sequence $(\varepsilon_n)_{n \in \mathbb{N}}$ of positive numbers. By assumption, for every $n$ there are disjoint measurable $A_n, B_n$ with $E = A_n \cup B_n$ and $\mu(A_n) < \varepsilon_n$, $\nu(B_n) < \varepsilon_n$. Now let
$$A = \limsup_{n \to \infty} A_n\quad \text{and}\quad B = \limsup_{n \to \infty} B_n\,.$$
Then $A,B$ are measurable with $A \cup B = E$ (if $x \notin A$, there is an $N_x$ such that $x \notin A_n$ for all $n \geqslant N_x$, and hence $x \in B$), but in general we have $A \cap B \neq \varnothing$. That doesn't matter however. If $\nu(B) = 0$, then also $\nu(E\setminus A) = 0$ since $E\setminus A \subset B$, so we can take $E_1 = A$ and $E_2 = E \setminus A$.
It remains to see that we can choose $(\varepsilon_n)$ in such a way that $\mu(A) = 0 = \nu(B)$. Sufficient for that is
$$\sum_{n = 0}^{\infty} \varepsilon_n < +\infty\,.$$