Proving that a set is nowhere dense. Let $A\subset X$ be dense in $X$. If $E$ is closed in $X$ and $E\cap A = \emptyset$, then I want to prove that $E$ is nowhere dense.
My attempt:
We will prove that $X \setminus \overline{E}=X \setminus E$ is dense in $X$. So we first note that since $A$ is dense in $X$ we have that for all $\epsilon >0$ and $x \in X$, $B_{\epsilon}(x) \cap A \not= \emptyset$.
Now we will prove that $A \subset X\setminus E$, and this is beause if we have $x \in A$ but $x \notin X\setminus E$  this implies that $x \in E \Rightarrow E\cap A \not= \emptyset $, so $A \subset X \setminus E$, but the first part of the proof imply that $\forall \epsilon >0 $ and $x \in X$ $B_{\epsilon}(x) \cap X \setminus E  \not= \emptyset$ and this means that $X \setminus E$ is dense and therefore $E$ is nowhere dense.
Then my question is, Am I right in my proof? or what do I have to fix or change.
My definition of nowhere dense:
A set $A$ is nowhere dense If the set $X \setminus \overline{A}$ is dense.
Thanks a lot in advance. 
 A: Your proof that $X\setminus E$ is dense in $X$ is correct (or at least seems correct to me), although it could use a bit more spacing and punctuation for the sake of the reader. 
Now, the last step, when you say "$X\setminus E$ is dense, hence $E$ is not dense", may require a bit more care and justification -- the statement is false.... $\mathbb{Q}$ is dense in $\mathbb{R}$, but so is $\mathbb{R}\setminus\mathbb{Q}$. And you're asked to prove nowhere density, not "not density."
Edit: Following the clarification in the question and the subsequent discussion in the comments below (copied here):

OK, [so] your definition is equivalent to the standard one. Anyway, for your proof to be "correct", replace in the last sentence "is not dense" by "is nowhere dense" (otherwise, your conclusion does not match what you aim at proving): "$X\setminus E$ is dense, and therefore $X\setminus \bar{E}$ is dense as $E=\bar{E}$: by definition, this means $E$ is nowhere dense."

A: Try this way, let $x\in E$ be interior point. then $B(x,r)$ for some $r$ positive,  is contained in $E$ and hence open and hence must intersect $A$ which is contradiction
A: Since E is closed, it is nowhere dense if its interior is empty. If A doesn't intersect E, so E is in the complement of A (which has empty interior: the complement of a dense set has no interiro). So E is insid a set with no interior, it has no interior.
A: There is a serious mistake in your reasoning, i.e. assuming that $E$ is nowhere dense if $X\setminus E$ is dense. The counter example to that claim is that both $\mathbb Q^c$ and $\mathbb Q$, the irrational and rational, are dense in $\mathbb R$.
A: Let $U$ denote the interior of the closure of $E$. 
Since $E$ is
closed it coincides with its closure so $U$ is also the interior
of $E$. 
Then $U$ is an open subset of $E$. 
If $U\neq\varnothing$
then $A\cap U\neq\varnothing$ since $A$ is dense and $U$ is open. 
This contradicts
$A\cap E=\varnothing$ since $U\subseteq E$, so we conclude that $U=\varnothing$. 
Proved
is now that $E$ is nowhere dense (i.e. the interior of its closure
is empty).
A: If E is closed, it is nowhere dense if its interior is empty. Now, since E doesn't intersect A, so it is in the complement of A, which has no interior (the complement of a dense set has no interior and vice versa). If E is inside a set with no interior, it has no interior.
