[proofplan]
Fix two points $x,y \in X$. The triangle inequality compares $d(x,a)$ and $d(y,a)$ for every point $a \in A$, and taking infima over $A$ turns this pointwise comparison into an inequality between $\operatorname{dist}_A(x)$ and $\operatorname{dist}_A(y)$. Repeating the same argument with $x$ and $y$ exchanged gives the opposite one-sided estimate. The two one-sided estimates are exactly the absolute-value Lipschitz bound.
[/proofplan]
[step:Verify that the distance-to-set values are finite real numbers]
Because $A$ is nonempty, choose a point $a_0 \in A$. For every $z \in X$, the set
\begin{align*}
S_z := \{d(z,a) : a \in A\}
\end{align*}
is a nonempty subset of $[0,\infty) \subset \mathbb{R}$, so it is bounded below by $0$. Since $d(z,a_0) \in S_z$, its infimum satisfies
\begin{align*}
0 \leq \inf S_z \leq d(z,a_0) < \infty.
\end{align*}
Hence $\inf S_z$ is a finite real number. Therefore $\operatorname{dist}_A(z) \in \mathbb{R}$ for every $z \in X$, so the function $\operatorname{dist}_A: X \to \mathbb{R}$ is well-defined.
[/step]
[step:Use the triangle inequality to compare the two infima]
Fix $x,y \in X$. For every $a \in A$, the triangle inequality in the [metric space](/page/Metric%20Space) $(X,d)$ gives
\begin{align*}
d(x,a) \leq d(x,y)+d(y,a).
\end{align*}
Since $d(x,y)$ is independent of $a$, taking the infimum over $a \in A$ on both sides gives
\begin{align*}
\inf_{a \in A} d(x,a) \leq \inf_{a \in A} \bigl(d(x,y)+d(y,a)\bigr).
\end{align*}
For a fixed real constant $c := d(x,y)$ and the nonempty set $S_y := \{d(y,a) : a \in A\}$, the elementary translation rule for infima gives
\begin{align*}
\inf_{a \in A} \bigl(d(x,y)+d(y,a)\bigr)=d(x,y)+\inf_{a \in A} d(y,a).
\end{align*}
Using the definition of $\operatorname{dist}_A$, we obtain
\begin{align*}
\operatorname{dist}_A(x) \leq d(x,y)+\operatorname{dist}_A(y).
\end{align*}
Equivalently,
\begin{align*}
\operatorname{dist}_A(x)-\operatorname{dist}_A(y) \leq d(x,y).
\end{align*}
[guided]
Fix $x,y \in X$. We want to compare the two numbers $\operatorname{dist}_A(x)$ and $\operatorname{dist}_A(y)$, but each is defined as an infimum over all points of $A$. The way to compare such infima is first to compare the quantities inside the infimum point by point.
Let $a \in A$ be arbitrary. Since $d$ is a metric on $X$, the triangle inequality applied to the triple $x,y,a \in X$ gives
\begin{align*}
d(x,a) \leq d(x,y)+d(y,a).
\end{align*}
This inequality holds for every $a \in A$. Therefore the set of values $\{d(x,a):a \in A\}$ is pointwise bounded above by the corresponding translated set $\{d(x,y)+d(y,a):a \in A\}$. By the monotonicity property of the infimum, taking infima over the same nonempty index set $A$ preserves the inequality:
\begin{align*}
\inf_{a \in A} d(x,a) \leq \inf_{a \in A} \bigl(d(x,y)+d(y,a)\bigr).
\end{align*}
Now observe that $d(x,y)$ does not depend on $a$. Define the real constant $c := d(x,y)$ and the nonempty set
\begin{align*}
S_y := \{d(y,a):a \in A\}.
\end{align*}
Because adding a fixed real constant translates every element of a nonempty bounded-below set, the infimum translates by the same constant:
\begin{align*}
\inf_{a \in A} \bigl(d(x,y)+d(y,a)\bigr)=d(x,y)+\inf_{a \in A}d(y,a).
\end{align*}
Substituting the definition of the distance-to-set function gives
\begin{align*}
\operatorname{dist}_A(x) \leq d(x,y)+\operatorname{dist}_A(y).
\end{align*}
Finally, subtracting the finite real number $\operatorname{dist}_A(y)$ from both sides yields the one-sided estimate
\begin{align*}
\operatorname{dist}_A(x)-\operatorname{dist}_A(y) \leq d(x,y).
\end{align*}
[/guided]
[/step]
[step:Exchange the two points to obtain the opposite one-sided bound]
Applying the previous argument with $x$ and $y$ interchanged gives
\begin{align*}
\operatorname{dist}_A(y)-\operatorname{dist}_A(x) \leq d(y,x).
\end{align*}
Since $d$ is symmetric, $d(y,x)=d(x,y)$, and hence
\begin{align*}
\operatorname{dist}_A(y)-\operatorname{dist}_A(x) \leq d(x,y).
\end{align*}
[/step]
[step:Combine the one-sided estimates into the Lipschitz inequality]
The two estimates just proved are
\begin{align*}
\operatorname{dist}_A(x)-\operatorname{dist}_A(y) \leq d(x,y)
\end{align*}
and
\begin{align*}
-\bigl(\operatorname{dist}_A(x)-\operatorname{dist}_A(y)\bigr) = \operatorname{dist}_A(y)-\operatorname{dist}_A(x) \leq d(x,y).
\end{align*}
For any real number $r$, the pair of inequalities $r \leq c$ and $-r \leq c$ with $c \geq 0$ is equivalent to $|r| \leq c$. Applying this with
\begin{align*}
r := \operatorname{dist}_A(x)-\operatorname{dist}_A(y), \qquad c := d(x,y),
\end{align*}
we obtain
\begin{align*}
|\operatorname{dist}_A(x)-\operatorname{dist}_A(y)| \leq d(x,y).
\end{align*}
Since $x,y \in X$ were arbitrary, $\operatorname{dist}_A:X \to \mathbb{R}$ is $1$-Lipschitz.
[/step]
