Confusion with closure of sets

Question

In one of the solutions to the problem of open sets gives this line,

$\overline{X-U} \cap U \subset \overline{(X-U)\cap U}=\emptyset$

$\implies \overline{X-U} \subset X-U$

$\implies X-U \space\text{is closed}$

My question is why is $X-U$ closed? (Note the overline here indicates the closure of the overlined set).

EDIT: Second Question:

If $A$ is a dense subset of $X$ and $U $ is open, then $A\cap U$ is dense in $U$. So in the solution it says, $x\in U \implies x\in X=\overline{A}$, my question is, why is $x \in U$ also, why does $x \in U$ imply that $x \in X?$

Answer 1

The closure is the smallest closed set which contains $X-U$, so if $\overline{X-U} \subset X-U$, then $\overline{X-U} = X-U$ which means that $X-U$ is closed

Answer 2

For $\overline{A}\subseteq A$, then $A$ is closed. One can prove that $A^{c}$ is open: For $a\in A^{c}$, since $a\notin\overline{A}$, there exists some open set $G$ that containing $a$ but $G\cap A=\emptyset$, so $a\in G\subseteq A^{c}$, we are done.

Definition of closure: $a\in\overline{A}$ if for every open set $G$ that containing $a$, then $G\cap A\ne\emptyset$.