$X \setminus A$ is dense iff for every nonempty open set $U$, $U \setminus A$ is nonempty For the first direction I assume that $X \setminus A$ is dense so that $\mathfrak{cl}(X \setminus A)=X$, let $U$ be a nonempty set if $U \setminus A$ is empty then $U$ is subset from $A$ hence $X \setminus A$ subset from $X \setminus U$ but $X \setminus A$ is subset from $\mathfrak{cl}(X \setminus A)$ so $\mathfrak{cl}(X \setminus A)$ is subset from $X \setminus A$ which contradicts that $X \setminus A$ is dense in $X$ ...
Is this proof correct?
How about the other direction because I worked on it but I can't reach the result ..
 A: Your proof is correct: If $U$ is open, with $U \setminus A$ empty, then $A$ contains $U$, so $X \setminus U$ contains $X \setminus A$. But $X \setminus U$ is a closed set, so it contains the closure of $X \setminus A$, too. But we assumed $X \setminus A$ was dense, so this closure is all of $X$. Therefore $X \setminus U$ is all of $X$, so $U$ is empty. 
Conversely, suppose that for all non-empty open sets $U$, the set $U \setminus A$ is non-empty. Then no non-empty subset of $A$ is open. Taking complements, no superset of $X \setminus A$ (except for $X \setminus \emptyset = X$) is closed. Therefore, by definition,
$$Cl(X \setminus A) = \bigcap_{F \supset X \setminus A, \\\text{F closed}}F = X$$
as required.
A: Your proof is correct, but it can be made simpler (without using contradiction).
If you set $B=X\setminus A$, then $U\setminus A=U\cap(X\setminus A)=U\cap B$. With this change your statement becomes

$B$ is dense if and only if, for every nonempty open set $U$, $U\cap B$ is not empty.

Suppose $B$ is dense and that $U$ is a nonempty open set and take $x\in U$. Since $B$ is dense, $x\in\operatorname{cl}(B)$, so for every neighborhood $V$ of $x$, $V\cap B\ne\emptyset$. But $U$ is a neighborhood of $x$, so $U\cap B\ne\emptyset$.
Suppose $B$ is not dense. Then there is $x\notin\operatorname{cl}(B)$ and, by definition, there is an open neighborhood $U$ of $x$ (so $U$ is open and nonempty) such that $U\cap B=\emptyset$.
