(X,$\tau$) is a topological space. E $\subseteq X$. If E is dense in X and X \ E is dense in X, then there are no isolated points in E. I proceed by contradiction. Then, there exists a point which is isolated. That means that there exists a neighborhood U of that point such that:
$$ U \cap E = \{p\} $$
This means that p $\in$ E. Thus, p is in the closure of E.
But the closure of E equals the closure of $X\setminus E$ because they are both dense in X. If p is in the closure of $X\setminus E$ then for all neighborhood of p we can write that:
$$ V \cap ( X \setminus E ) \neq \emptyset $$
This means that: $$ (V \cap X ) \setminus  E \neq  \emptyset $$
This means that p $\notin$E. But this is a contradiction. 
Is this proof right? I always get to a conclusion but I always struggle to understand if everything that I did is acceptable. 
Thank you! 
 A: Assuming $X$ is Hausdorff (but $T_1$ would be sufficient).
Let $p\in E$ and let $U$ be an open neighborhood of $p$ (in $X$) with $U\cap E=\{p\}$. By assumption, $U\cap(X\setminus E)\ne\emptyset$.
Let $q\in U\cap(X\setminus E)$; then $U'=U\setminus\{p\}$ is an open neighborhood of $q$, so $U'\cap E\ne\emptyset$: a contradiction.
If the space is not $T_1$ the statement is false: if $X=\{0,1\}$ has the indiscrete topology, both $\{0\}$ and $\{1\}$ are dense.
A: Your proof is wrong and here is the problem: you get to $(V\cap X)\setminus E\neq\emptyset$ for all open neghbourhoods $V$ of $p$. But, $V\cap X = V$, so what you have is that $V\setminus E\neq \emptyset$. This doesn't actually tell you that $p\not\in E$. Why would it? All this tells you is that there are no open neighbourhoods of $p$ completely contained in $E$, which just tells you that $E$ is not open in $X$.
egreg already posted a counterexample to the claim as is. However, it is true that there are no isolated points in $X$. Assume that $p$ is isolated point in $X$. It means that $\{p\}$ is open in $X$. However, if $p\in E$, then there exists open set in $X$ not intersecting $X\setminus E$ (namely $\{p\}$), so $X\setminus E$ is not dense in $X$, and if $p\not\in E$, the same argument says that $E$ is not dense in $X$.
