In a metric space with no isolated points, why is the closure of a discrete set nowhere dense? Does anyone know why this is true?  I know that the closure of a set in a metric space is the set of points a distance of 0 away from the original set.  Also, a discrete set is one with a neighborhood about any point whose union with the discrete set is that point.

Let $d\colon X \times X \to \mathbf R$ be the metric, and for $x \in X$ and $r > 0$ define
$$
B(x, r) = \{y \in X : d(x, y) < r\}
$$
as usual. Writing $D$ for this discrete subset, suppose we have a non-empty open set $U \subset \overline{D}$. There must be an $a \in U \cap D$; find an $r > 0$ such that $B(a, r)$ is contained in $U$ and contains no other points of $D$. Since there are no isolated points, there is a $y \neq a$ in $B(a, r)$. Can you find a ball around $y$ that doesn't intersect $D$?
