Let $X$ be a metric space without isolated points. Then the closure of a discrete set in $X$ is nowhere dense in $X$. I am having some trouble with this question from Willard's General Topology (p.37). I found that this was asked previously 9 years ago (see link: In a metric space with no isolated points, why is the closure of a discrete set nowhere dense?) but I'm having trouble following the hint at the end of the argument in the top rated answer.
I tried answering but the user hasn't been online since 2014, so I hope it's okay to re-post and hopefully get a more fully fleshed out answer.
The user said:

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 $D$?

I'm also not clear on why $U \subseteq \overline{D}$ being non-empty and open means that $\exists a \in U \cap D$. Obviously, $\exists a \in U \cap \overline{D} \implies (a \in U \text{ and } a \in \overline{D})$. But how do we get that $a \in D$ rather than, say, $a \in \text{Boundary}(D)$.
And again, I'm not clear on where to go from the hint at the end.
Thank you.
 A: 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 $y \in U_d\setminus \{d\}$, 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
A: The point is that if $B$ is an open ball around $y$ that does not intersect $D$, then $y\notin\operatorname{cl}D$, which is impossible, since $y\in B(a,r)\subseteq U\subseteq\operatorname{cl}D$.
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 $\epsilon\le r-d(a,r)$, $a\notin B(y,\epsilon)$, since $\epsilon\le d(y,a)$, 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 $a\in U\cap D$, any non-empty open set that intersects $\operatorname{cl}D$ necessarily intersects $D$ by the definition of closure.
A: 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 $y \in \overline D$].
