is this correct? : let $C, A \subseteq X$ then
$X= A \cup X\setminus A= \overline{A} \cup \overline{X\setminus A}$ since both sets are closed and open trivially. Hence $C=X \cap C=(\overline{A} \cup \overline{X\setminus A})\cap C= (C\cap \overline{A})\cup (C\cap \overline{X\setminus A})$
$X= A \cup X\setminus A= \overline{A} \cup \overline{X\setminus A}$ is correct. We only need set theory to prove that. First equality is obvious. Since $A\subseteq \overline{A}$ and $X\setminus A\subseteq \overline{X\setminus A}$, we have $X=A\cup X\setminus A\subseteq \overline{A}\cup \overline{X\setminus A}$. Thus $X= \overline{A}\cup \overline{X\setminus A}$. Your approach of open and closed set is not only incorrect but don’t make sense.
