TRUE or FALSE:
Let $L_1, L_2$ be any two regular languages over the same alphabet $\Sigma$, then the language $L=\{w\in\Sigma^* \mid w\in L_1 \text{ or } w\notin L_2\}$ is regular.
So we have to determine whether $L_1 \cup \overline{L_2}$ is regular or not.
Proof:
First attempt:
First we have to prove that a complement of a regular language is also regular: the complement of a language $L$ with respect to an alphabet $\Sigma$ such that $\Sigma^*$ is $\Sigma^*-L$. Since $\Sigma^*$ is surely regular the complement of a regular language is always regular.
Let's prove that the union of two regular languages is also regular: For example, let $\Sigma = \{a,b\}$. Assume $L_1 = \{a\}$ and $L_2 = \{b\}$ so they are regular language. Then the union: $\{a\} \cup \{b\} = \{ab\}$ is also regular. Since $\{a\}$ is regular , $\{a\}^*$ is also a regular language.
After these two proofs we can say that the statement above is True.
Second attempt:
A regular language is regular iff there is a finite state machine recognizing it. Let $L_1 = \{S_1,\Sigma,\delta_1,s_0^1, F_1\}$ and $L_2 = \{S_2,\Sigma,\delta_2,s_0^2, F_2\}$ be two automata. First we have to take the complement of $L_2$. The complement of $L_2$ is the set of states without the set of final states: $\overline{L_2} = \{S_2,\Sigma,\delta_2,s_0^2, S_2-F_2\}$. Then we can create the product automaton of the two languages which is: $L = \{S_1 \times S_2,\Sigma,\delta_1 \times \delta_2,s_0^1 \times s_0^2, F_1\ \times (S_2-F_2)\}$. The final states of the language $L = L_1 \cup \overline{L_2}$ is the set of states where $F_1$ or $S_2 - F_2$ is final. Since there exists a finite state machine that recognizes the language $L$, we can say that $L$ is a regular language.
Can somebody please correct me? Maybe I made a mistake.