Proof of Vanishing of the Distance to a Set Characterizes the Closure

Let $S_{x,A}$ be the set of real numbers of the form $d(x,a)$ with $a\in A$, as in the definition of the distance to a set, so that $\operatorname{dist}_d(x,A)$ is the greatest lower bound of $S_{x,A}$; the set $S_{x,A}$ is nonempty and bounded below by $0$. Write $s<t$ to mean $s\le t$ and $s\ne t$.

Necessity of membership in the closure. Assume $x\in\operatorname{cl}_X(A)$. By claim 1 of The Distance to a Set is Nonexpansive we have $0\le\operatorname{dist}_d(x,A)$. Suppose $\operatorname{dist}_d(x,A)\neq 0$; then $0<\operatorname{dist}_d(x,A)$. Applying condition 3 of Characterization of the Closure in a Metric Space by Open Balls with $\varepsilon=\operatorname{dist}_d(x,A)$ produces $a\in A$ with $d(x,a)<\operatorname{dist}_d(x,A)$. But claim 2 of The Distance to a Set is Nonexpansive gives $\operatorname{dist}_d(x,A)\le d(x,a)$, so claim 2 of Elementary Order Arithmetic in an Ordered Field yields $\operatorname{dist}_d(x,A)<\operatorname{dist}_d(x,A)$, contradicting the requirement that the two sides of a strict inequality be distinct. Hence $\operatorname{dist}_d(x,A)=0$.

Sufficiency. Assume $\operatorname{dist}_d(x,A)=0$, and let $\varepsilon$ be a real number with $0<\varepsilon$. Then $\operatorname{dist}_d(x,A)<\varepsilon$, so claim 2 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$, applied to the nonempty set $S_{x,A}$ bounded below, produces $t\in S_{x,A}$ with $t<\varepsilon$. By the description of $S_{x,A}$ there is $a\in A$ with $t=d(x,a)$, so $d(x,a)<\varepsilon$. Since $\varepsilon$ was arbitrary, condition 3 of Characterization of the Closure in a Metric Space by Open Balls holds, and that theorem gives $x\in\operatorname{cl}_X(A)$.