The one-point set is nowhere dense in $X$ I am trying to show for a metric space $X$, a set $\{x\}$ consisting of a single point is nowhere dense. I have proven it by showing $[{\{x\}}]^o = \emptyset$ where $[A]$ is the closure of the set $A$ and $A^o$ is the interior of $A$. however I am struggling to understand the following proof:
"Choose an open ball $B$ $x \in B$, so that $B(x,r) \subset X$ for some $r$. Since $x$ is an accumulation point of $X$, there is a point $y \in B(x,r), y \not=x$. Since $B(x,r)$ is open we have $B(y,r_1) \subset B(x,r)$ for some $r_1 >0$. Furthermore for $r_0 = \min\{d(x,y),r_1\}$ we have $x \notin B(y,r_0) \subset B(x,r) \subset B$ which implies the result."
I don't understand why $x$ is an accumulation point of $X$, and why the conclusion means it's nowhere dense.
 A: Being nowhere dense depends on the ambient space (and his induced topology) as well as many topological properties. So the assertion "$x$ is an accumulation point of $X$" is a priori false. Consider $\mathbb{Z}$ , the set of integers. Seeing it as metric space , every point belonging to $\mathbb{Z}$ is isolated (and hence is not an accumulation point). But seeing it as subspace of the larger metric space  $\mathbb{R}$ is not longer true that each $x \in \mathbb{Z}$ is isolated. This responses your first question.
The second question is related  with the ambient space again : A set $A$ may be nowhere dense when considered as a subspace of a topological space $X$ but not when considered as a subspace of another topological space $Y$. For example, a line is nowhere dense in $\mathbb{R}^{2}$ , but is not on itself. (If fact every set is dense on itself). Thus, a single point may be not nowhere dense if is it isolated in the ambient space where its belong. The true fact is that "If  $X$ has no isolated points, then every finite set is nowhere dense''. The way the answer proves that the singleton $ \{x\} $ is nowhere dense , is using the fallowing equivalent proposition, only valid in metric spaces : 
$A$ is nowhere dense in the m. e. $X$ iff for every open ball $B$ there is a open ball $V \subseteq B $ such that $ V \cap A = \emptyset$
(For topological spaces in general,  the correct assertion is : A set $A$ is nowhere dense iff for every non-empty open set $U$ there is a non-empty open set $V$ such that $V\subseteq U$ and $A\cap V=\emptyset$ )
