Let $X$ be a metric space without isolated points. Then the closure of a discrete set in $X$ is nowhere dense in $X$.

I propose a simpler proof. It follows from the following folkore

Fact: Let $X$ be crowded (this means $X$ is $T_1$ so points are closed (this applies to all metric spaces, but we need no more of the metric than this) and $X$ has no isolated points). If $D subseteq X$ is discrete (so discrete in its subspace topology), then $D$ is open in $overline{D}$ and $overline{D}$ is nowhere dense.

Proof: for each $d in D$ we have an open subset $U_d$ of $X$ that obeys $U_d cap D = {d}$; this is just a restatement that $D$ is discrete. Then I claim that in fact $overline{D} cap U_d = {d}$ as well. (If $y neq d$ were in $overline{D} cap U_d$ then $y in U_dsetminus {d}$, which is open as $X$ is $T_1$, and $(U_d setminus {d}) cap D = {d} setminus {d} = emptyset$, contradicting $y in overline{D}$; so $overline{D} cap U_d = {d}$, as claimed). So ${d}$ is open in $overline{D}$ for each $d in D$ so $D$ is open in $overline{D}$ (union of open singletons).

To see that $overline{D}$ is nowhere dense, we only need to show it has empty interior in $X$, so suppose (for a contradiction) $O$ is non-empty open in $X$ such that $O subseteq overline{D}$. Then this $O$ intersects $D$ (as $D$ is dense in $overline{D}$ obviously), say in $d in O cap D$ and then $U_d cap O = U_d cap (O cap overline{D}) = (U_d cap overline{D}) cap O = {d}$ by the previous paragraph, but this contradicts that $X$ has no isolated points (note that $O cap U_d$ is open in $X$ and so ${d}$ would be). This shows that indeed the closure of $D$ is nowhere dense. QED

Take any $x in U$. Then $x in overline D$. Since $U$ is an open set containing $x$ and $x in overline D$ it follows that $U$ must intersect $D$, so there is a point $a$ in $U cap D$.

Since $D$ is discrete $a$ is not a limit point of $D$. So we can find a ball $B(a,r)$ which is contained in $U$ and contains no other points of $D$.

Finally take any ball around $y$ with radius less than $d(y,r)$ which is contained in $B(a,r)$. That will lead to a contradiction, right? [Recall that $y in overline D$].

The point is that if $B$ is an open ball around $y$ that does not intersect $D$, then $ynotinoperatorname{cl}D$, which is impossible, since $yin B(a,r)subseteq Usubseteqoperatorname{cl}D$.

That $B$ exists is easy: take $epsilon=min{d(a,r),r-d(a,r)}$ and use $B(y,epsilon)$. $B(y,epsilon)subseteq B(a,r)$, since $epsilonle r-d(a,r)$, $anotin B(y,epsilon)$, since $epsilonle d(y,a)$, and $B(y,epsilon)cap D=varnothing$, since $a$ is the only point of $D$ in $B(a,r)$.

As for why there is some $ain Ucap D$, any non-empty open set that intersects $operatorname{cl}D$ necessarily intersects $D$ by the definition of closure.