Are supersets of non-regular languages also non-regular? I have to proof that if $L_1 \subset L_2$ and $L_1$ is not regular then $L_2$ it not regular. This is my proof. Is it valid? 

Since $L_1$ is not regular, there does not exists a finite automata $M_1$ such that $L_1$ is the language of $M_1$. Pick $x\in L_1$. So $x \in L_2$ and suppose that $L_2$ is regular. Then there exists a finite automata $M_2$ such that $L_2$ is the language of $M_2$. Since $x \in L_2$ and $L_2$ is regular, there exists a state $s\in S$ such that from the initial state in $M_2$ there is a path $x$ to this final state $s$. Since this holds for all $x \in L_1$, we can construct a finite automata which language is $L_1$, so $L_1$ is regular, so we reached a contradiction, so $L_2$ is not regular.

Can this be done easier?
Since this holds for all $x\in L_1$, we can construct a finite automata which language is $L_1$.


And how would you do this? You have an infinite amount of paths in hands which you have to mold into a finite set of states and edges.

In fact, the statement does not hold.


   Remember that $\Sigma^*$ is regular but there are non-regular languages.