[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*}[/step]

[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: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]