Semi-decidability of the language $\overline{L_{\epsilon}}$
Assume for contradiction $\overline{L_{\epsilon}}$ is semi-decidable. It's easily proven that $L_{\epsilon}$ is semi-decidable by reduction to the HALTING problem, i.e. $L_{H}$ is semi-decidable. If $L_{\epsilon}$ and $\overline{L_{\epsilon}}$ are semi-decidable, hence $L_{\epsilon}$ is decidable. This is a contradiction, as $L_{\epsilon}$ as well as $L_{H}$ are semi-decidable.
In this prove the following claim was used:

$\textbf{Claim:} \text{ If $L$ and $\overline{L}$ are semi-decidable, then $L$ is decidable}$