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?