Proof of The Distance to a Set is Nonexpansive

For $z\in X$ let $S_{z,A}$ be the set of real numbers of the form $d(z,a)$ with $a\in A$, as in the definition of the distance to a set, so that $\operatorname{dist}_d(z,A)$ is the greatest lower bound of $S_{z,A}$; this set is nonempty and bounded below, so the greatest lower bound exists by Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below.

Claim 1. By condition 1 of the definition of a metric, $0\le t$ for every $t\in S_{x,A}$, so $0$ is a lower bound for $S_{x,A}$. A greatest lower bound dominates every lower bound, so $0\le\operatorname{dist}_d(x,A)$.

Claim 2. For $a\in A$ we have $d(x,a)\in S_{x,A}$, and a greatest lower bound is in particular a lower bound, so $\operatorname{dist}_d(x,A)\le d(x,a)$.

Claim 3. Let $a\in A$. By condition 4 of the definition of a metric, $d(x,a)\le d(x,y)+d(y,a)$, and by claim 2 we have $\operatorname{dist}_d(x,A)\le d(x,a)$; hence $\operatorname{dist}_d(x,A)\le d(x,y)+d(y,a)$ by transitivity of the order. Adding $-d(x,y)$ to both sides, which preserves $\le$ by the compatibility of the order with addition in an ordered field, and simplifying by commutativity and associativity of addition together with claim 3 of Additive Cancellation and Elementary Additive Identities in a Field, gives

Since $a\in A$ was arbitrary, $\operatorname{dist}_d(x,A)-d(x,y)$ is a lower bound for $S_{y,A}$, so it is at most the greatest lower bound:

Adding $d(x,y)$ to both sides and simplifying in the same way gives claim 3.

Claim 4. Adding $-\operatorname{dist}_d(y,A)$ to both sides of claim 3 and simplifying gives

Exchanging the roles of $x$ and $y$ in claim 3 and arguing in the same way gives $\operatorname{dist}_d(y,A)-\operatorname{dist}_d(x,A)\le d(y,x)$, and $d(y,x)=d(x,y)$ by condition 3 of the definition of a metric. Write $u=\operatorname{dist}_d(x,A)-\operatorname{dist}_d(y,A)$; by claim 6 of Additive Cancellation and Elementary Additive Identities in a Field the second inequality reads $-u\le d(x,y)$, which by claim 4 of Elementary Order Arithmetic in an Ordered Field is equivalent to $-d(x,y)\le u$. So $-d(x,y)\le u$ and $u\le d(x,y)$, and claim 6 of Properties of the Absolute Value in an Ordered Field gives $|u|\le d(x,y)$, which is claim 4.

Claim 5. Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$, and take $\delta=\varepsilon$, so that $0<\delta$. Let $x,y\in X$ satisfy $d(x,y)<\delta$. By the definition of $d_{\mathbb{R}}$ and by claim 4,

so $d_{\mathbb{R}}(f(x),f(y))<\varepsilon$ by claim 2 of Elementary Order Arithmetic in an Ordered Field. This is exactly uniform continuity of $f$ on $X$.

For continuity on $X$, let $x\in X$ and let $\varepsilon>0$; take the same $\delta=\varepsilon$. Every $y\in X$ with $d(x,y)<\delta$ satisfies $d_{\mathbb{R}}(f(x),f(y))<\varepsilon$ by the previous paragraph, and $d_{\mathbb{R}}(f(y),f(x))=d_{\mathbb{R}}(f(x),f(y))$ by condition 3 of the definition of a metric applied to $d_{\mathbb{R}}$, which is a metric by The Absolute Value Metric on the Real Line. Hence $f$ is continuous at $x$ relative to $X$, and as $x$ was arbitrary, continuous on $X$.