You don't need to use $\varepsilon$ to see that $\mu_{\sup}\geq\max\{\mu_1,\mu_2\}$. Since $\emptyset$ and $X$ are elements of $\Sigma$, for any $E\in\Sigma$,
$\mu_\sup(E)=\sup_{F\in\Sigma}\,\{\mu_1(E\cap F)+\mu_2(E\setminus F)\}\geq \left\{\begin{align} &\mu_1(E\cap \emptyset)+\mu_2(E\setminus \emptyset)=\mu_2(E)\\ &\mu_1(E\cap X)+\mu_2(E\setminus X)=\mu_1(E) \end{align}\right.\quad. $
Now let $\mu$ be a nonnegative measure defined on $\Sigma$, such that, for all $E\in\Sigma$, we have $\mu(E)\geq \max\{\mu_1(E),\mu_2(E)\}$. Let $E\in\Sigma$ and $F\in\Sigma$. Then, we have
$\mu_1(E\cap F)+\mu_2(E\setminus F)\leq \mu(E\cap F)+\mu(E\setminus F)=\mu(E)$
and taking the supremum over $F\in\Sigma$ we get $\mu_{\sup}(E)\leq \mu(E)$.
What you showed with the $\varepsilon$ is that $\mu_{\sup}\leq \mu_1+\mu_2$, but the reversal inequality doesn't need to be true (to see that, take $\mu_1=\mu_2$ and $E\in\Sigma$ such that $\mu_1(E)\neq 0$).