Two regular languages over the same alphabet, regular or not regular?
I see several problems.

In your first part you say that, given a regular language $L$, $\overline{L}$ is regular since  $\Sigma^*- L$ is also regular. This is true, but it is implicitly using the fact that regular languages are closed under difference. It is not clear if you're allowed to use that fact as a black box (since it is a more general statement than the one you're trying to prove).

In your second part you are not giving a proof. You are just showing that $L_1 \cup L_2$ is regular when $\Sigma=\{a,b\}$, $L_1 = \{a\}$ and $L_2 = \{ b \}$. What if $\Sigma \neq \{a,b\}$? What if $L_1 \neq \{a \}$?
Your proof needs to work for all possible choices of $\Sigma, L_1,$
 and $L_2$.

Moreover, $L_1 \cup L_2 \neq \{ab\}$ contrarily to what you claim.

Finally, you say that since $\{a\}$ is regular then so is $\{a\}^*$. This is true but it is unclear why you need this.