Proof set is nowhere dense iff the complement of the closure is dense Suppose $(S,d)$ is a metric space and $E\subset S$. Prove that $E$ is nowhere dense if and only if $S\setminus \bar{E}$ is dense.
I have yet been able to prove the $\Leftarrow$ part. Suppose $E$ is not nowhere dense, this implies that $\bar{E}^\circ\neq\emptyset$. So, $\exists x\in\bar{E}$ and $\exists\epsilon>0$ such that $B(x,\epsilon)\in\bar{E}$. But this contradicts that $S\setminus \bar{E}$ is dense. 
Now the $\Rightarrow$ part. Suppose $S\setminus\bar{E}$ is not dense, which implies that there is an non-empty open set $V$ such that $V\cap S\setminus\bar{E}=\emptyset$. This implies there exist $x\in V$ such that $x\in \bar{E}$, so there exists a sequence $(x_n)_{n \in\mathbb{N}}$ converging to $x$. 
How can I proceed from here?
 A: Although OP asked about metric spaces, let me answer it in a more general context. For a topological space $X$ and $B\subset X$, one has
$$B^\circ=X\setminus \overline{X\setminus B}.$$
Indeed, since $B^\circ\subset B$, it follows that $B^\circ\cap X\setminus B=\emptyset$, hence $B^\circ\subset X\setminus \overline{X\setminus B}$. Conversely, if $x\not\in \overline{X\setminus B}$, then there exists an open set $V$ such that $x\in V$ and $V\cap X\setminus B=\emptyset$, hence $x\in V\subset B$, i.e., $x\in B^\circ$.
Now, $E\subset X$ is nowhere dense iff $\overline{E}^\circ =\emptyset$, but
$$\overline{E}^\circ=\emptyset\Leftrightarrow X\setminus\overline{X\setminus \overline{E}}=\emptyset\Leftrightarrow \overline{X\setminus \overline{E}}=X,$$
meaning that $X\setminus \overline{E}$ is dense.
A: Indeed, if $S \setminus \overline E$ is not dense, then we find an open $V$ such that $V \cap S \setminus \overline E = \emptyset$. Now
$$
V \cap S \setminus \overline E = S \setminus (\overline E \cup (S \setminus V))
$$
and we get $\overline E \cup (S \setminus V) = S$. Thus, $V \subseteq \overline E$.
