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. 

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 $\mathbb R$.
