Here $L_1 = \{\{\}\}$ is a set of sets and not a set of strings and hence is not a language.
Suppose $L_1 = \{ \epsilon\}$,
Now, all strings (only 1) in $L_1$ is present in $L_2$. Is there anyother string in $L_2$ - yes, if $\Sigma$ is non-empty. So, if $\Sigma$ is non empty, then $L_1$ is a proper subset of $L_2.$