Proving $S=SL\implies S=\emptyset$ Let $L\subseteq \Sigma^*$ such that $\{\epsilon\}\not\in L$. Then for any $S\subseteq \Sigma^*, 
 S=SL\implies S=\emptyset$.

So we suppose $S=SL$ and $S\ne\emptyset$. Then $\exists w\in S$ such that $0\le|w|\le |v|$ for some $v\in S$. Now $|w|\ne 0$, since $|w|=0\implies\{\epsilon\}=w\in S=SL\implies \{\epsilon\}=xy$ for some $x\in S$ and $y\in L\implies x=y=\{\epsilon\}\in L$ which contradicts that $\{\epsilon\}\not\in L$.
Thus $w\ne\{\epsilon\}$ and $\{\epsilon\}\not\in S$. So we have
\begin{align}
w \in S &amp;\implies w\in SL\\
&amp;\implies w=sl\;\text{for some}\; s\in S\; \text{and}\; l\in L\\
&amp;\implies |w|=|s|+|l|
\end{align}
Here I'm stuck now. I think if I can show that $|l|=0$, then we can arrive at a contradiction.
Your help is highly appreciated.
You should consider (one of) the word $w$ of minimal length in $S$, and show that for this word in particular, you can reach the conclusion that $|l| = 0$.