Is this proof right?
I approached this problem by showing that $B(x;\epsilon )$ is a subset of the complement of the closure of $V$ for all $x$ in $X$ and $\epsilon$ bigger than $0$. We know $B(x;\epsilon )= \{a \in X: d(x,a) < \epsilon \}$. Let $b \in B(x;\epsilon )$. Let us prove that $b$ is also in $(\overline{V})^c$. This would be the same as proving that $b$ is not in $\overline{V}$. Suppose $b$ would be in $\bar{V}$. This would mean that $B(x;\epsilon ) = \{b \in \overline{V}: d(x,b) < \epsilon \}$. Since $V$ is closed, $\overline{V}=V$, and thus $b$ would be in $V$. However, by the closedness of $V$, there does not exist an epsilon neighborhoud around $V$ such that $b$ would be in an epsilon distance from $x$ (I also realize now that $x$ can be in $V$, so I am not sure what to do then?), and thus $V$ is nowhere dense.
Could anyone give a hint or advice how to look at this? It will be appreciated a lot!