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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.

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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 $D$?

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If $D$ is the discrete set in question and $x$ is in $D$, let $U$ be an open neighborhood of $\{x\}$ disjoint from $D\setminus\{x\}$. Then $U$ is also disjoint from $\overline{D\setminus\{x\}}$. In particular, this means that $\{x\}$ is not in $\overline{D\setminus\{x\}}$, which implies that $\overline{D\setminus\{x\}}=\overline D\setminus\{x\}$. Thus $U$ is disjoint from $\overline D\setminus\{x\}$. Since $x$ is not an isolated point, this implies that $U$ is not contained in $\overline D$. Considering that $U$ could be taken to be any sufficiently small ball centered at $\{x\}$, this shows that $x$ is not in the interior of $\overline D$.

Suppose that $y$ is an element of $\overline D$, and $V$ is a neighborhood of $y$. Then there is an $x$ in $V\cap D$. Since $V$ is a neighborhood of $x$, $V$ is not contained in $\overline D$ by the previous paragraph. Therefore $y$ is not an interior point of $\overline D$. Since a closed set is nowhere dense if and only if it has empty interior, this shows that $\overline D$ is nowhere dense.

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