Theorem about dense and closed sets 
Let X a topological space. A subset $A \subset X$ is said to be dense in $X$, if $\overline{A}=X$. $A$ is said to be nowhere dense in $X$, if $(\overline{A})^{°}=\emptyset$. If $A\subset X$ is closed, then $A$ is nowhere dense in $X$, if $X\setminus A$ is dense in $X$. Is this theorem valid for any set?

I'm absolutely confused and disorientated to begin the argumentation. Do you have any hints, to start or think at?
Thank you! 
 A: Yes, a theorem valid in any space $X$ for any subset $A$ is:
$$X \setminus A^\circ = \overline{X \setminus A}$$ from which your fact follows immediately. 
($x$ is not in the interior of $A$ iff there is no open set containing $x$ that stays inside $A$ iff every open set that contains $x$ intersects $X \setminus A$ iff $x$ is in the closure of $X\setminus A$.)
A: Since $X \setminus A$ is dense in $X$,
$\overline{X \setminus A} = X$
$\Rightarrow X \setminus A^{\circ} = X$
$\Rightarrow A^{\circ} = \emptyset$
$\Rightarrow \overline{A}^{\circ} = \emptyset \quad$ since $A$ is closed, which implies that $A$ is nowhere dense in X.
Edit: We prove that $\overline{X \setminus A} = X \setminus A^{\circ}$ by showing that they are contained in each other:
Let $A'$ represent the set of limit points of $A$.
Consider $x \in \overline{X \setminus A} = X \setminus A \cup \left(X \setminus A \right)'$
If $x \in X \setminus A \subseteq X \setminus A^{\circ}$ (since $A^{\circ} \subseteq A$ always)
$\Rightarrow \overline{X \setminus A} \subseteq X \setminus A^{\circ}$
On the other hand, if $x \in \left(X \setminus A \right)'$, then $\forall \, \epsilon > 0$ we have:
$N_{\epsilon}(x) \cap \left(\left(X \setminus A \right)\setminus \{x\}\right) \neq \emptyset$
$\Rightarrow N_{\epsilon}(x)\not\subseteq \left( X \setminus \left(X \setminus A\right) \setminus \{x\} \right) = A \cup \{x\}$
$\Rightarrow N_{\epsilon}(x) \not\subseteq A$
So $x$ is not an interior point of $A$.
$\Rightarrow x \in X \setminus A^{\circ}$
$\Rightarrow \overline{X \setminus A} \subseteq X \setminus A^{\circ} $
Conversely, let $x \in  X \setminus A^{\circ}$
$\Rightarrow \forall \, \epsilon > 0 \, \, N_{\epsilon}(x) \not \subseteq A$
$\Rightarrow N_{\epsilon}(x) \cap X \setminus A \neq \emptyset $
$\Rightarrow x$ is a limit point of $A$ or a point of $A$ or both,
$\Rightarrow x \in  X \setminus A \cup \left(X \setminus A \right)' = \overline{X \setminus A}$
$\Rightarrow  X \setminus A^{\circ}\subseteq \overline{X \setminus A}$
Thus $X \setminus A^{\circ}= \overline{X \setminus A}$
