$A$ is dense iff there is not open subset nonempty in $X \setminus A$ $A$ is dense iff there is not open subset nonempty in $X \setminus A$. Let $X$ be topological space. 
 my try: If $A$ is dense then $X = Cl(A) $. we can assume there is an open $O$ such that $O \subseteq X \setminus A $. My question is : Can we assume that if $X = cl(A) \; \;then\; \; X = A $?? Im kind of stuck here. 
For the other direction, if there is not open subset nonempty in $X \setminus A$. we must show $X = Cl(A)$. In other wrods, we want to show that every $x \in X$ is a limit point. so pick $x \in X$ Take a neighborhood $N$ of $x$. We want to show $N \cap A \neq \varnothing$ if $N \cap A = \varnothing $, then $N \subseteq X \setminus A $. So we have an open set inside $X \setminus A$ which contradicts hypothesis. so $A$ must be dense.
Is this correct?
 A: You can't assume that $X = \operatorname{cl}(A)$ (take for example $X = \mathbb{R}$ and $A = \mathbb{Q}$). 
Here is a method to prove your result without discussing limits. Suppose that $A$ is dense and let $U \subset A^c$ be open. Then $A \subset U^c$ and $U^c$ is closed. Therefore $\operatorname{cl}(A) \subset U^c$ (since every closed set containing $A$ also contains $cl(A)$). But this implies that $U \subset \operatorname{cl}(A)^c = \varnothing$. So $U = \varnothing$. Conversely, suppose that the only open subset of $A^c$ is $\varnothing$. Let $K$ be any closed subset of $X$ with $A\subset K$. Then $K^c$ is an open set and $K^c \subset A^c$. By assumption then $K^c = \varnothing$, so $K = X$. Therefore, the only closed subset of $X$ containing $A$ is $X$, implying that $\operatorname{cl}(A) = X$, which proves that $A$ is dense.
A: For the first part, no, $X = \mathrm{cl} (A)$ does not imply $A = X$.  For an example, consider the rationals $\mathbb{Q}$ as a subset of the real line.  For this direction it might be easier to prove the contrapositive: if there is a nonempty open $U \subseteq X \setminus A$, then $A$ is not dense.  (Hint: Can any $x \in U$ belong to $\mathrm{cl}(A)$?  Prove it!)
The main line of your proof of the opposite direction seems fine.  (Just be sure to mention that you have a nonempty open set inside $X \setminus A$.)
