[proofplan] We compare the distance from $x$ to $M$ with the distance from $y$ to $M$ by inserting $y$ between $x$ and an arbitrary point $m \in M$. The triangle inequality gives an upper bound for $\|x-m\|_X$ in terms of $\|x-y\|_X$ and $\|y-m\|_X$, and taking the infimum over $m$ gives $E_M(x) \leq \|x-y\|_X + E_M(y)$. Repeating the same argument with $x$ and $y$ interchanged gives the opposite one-sided estimate, and the two estimates are exactly the desired absolute-value bound. [/proofplan]

[step:Compare $E_M(x)$ to $E_M(y)$ using the triangle inequality]Fix $x,y \in X$. Since $M$ is nonempty, the set $\{\|x-m\|_X : m \in M\}$ is nonempty for each $x \in X$, so $E_M(x)$ is well-defined as a finite nonnegative real number. Let $m \in M$ be arbitrary. By the triangle inequality in the normed space $X$, \begin{align*} \|x-m\|_X \leq \|x-y\|_X+\|y-m\|_X. \end{align*} Since $\|x-y\|_X$ is independent of $m$, taking the infimum over all $m \in M$ gives \begin{align*} E_M(x) = \inf_{m \in M}\|x-m\|_X \leq \inf_{m \in M}\bigl(\|x-y\|_X+\|y-m\|_X\bigr) = \|x-y\|_X+E_M(y). \end{align*}[/step]

[guided]Fix $x,y \in X$. The goal is to compare the two infima defining $E_M(x)$ and $E_M(y)$. Because $M$ is nonempty, for every point $z \in X$ the set $\{\|z-m\|_X : m \in M\}$ is nonempty and bounded below by $0$, so \begin{align*} E_M(z)=\inf_{m \in M}\|z-m\|_X \end{align*} is a well-defined element of $[0,\infty)$. Now choose an arbitrary point $m \in M$. The triangle inequality applied to the decomposition \begin{align*} x-m = (x-y)+(y-m) \end{align*} gives \begin{align*} \|x-m\|_X \leq \|x-y\|_X+\|y-m\|_X. \end{align*} This estimate holds for every $m \in M$. The term $\|x-y\|_X$ does not depend on $m$, so taking the infimum over $m \in M$ preserves the inequality and yields \begin{align*} \inf_{m \in M}\|x-m\|_X \leq \inf_{m \in M}\bigl(\|x-y\|_X+\|y-m\|_X\bigr). \end{align*} Since adding a fixed real number commutes with taking an infimum over a nonempty set, \begin{align*} \inf_{m \in M}\bigl(\|x-y\|_X+\|y-m\|_X\bigr)=\|x-y\|_X+\inf_{m \in M}\|y-m\|_X. \end{align*} Therefore \begin{align*} E_M(x) \leq \|x-y\|_X+E_M(y). \end{align*}[/guided]

[step:Reverse the comparison by interchanging $x$ and $y$] Applying the previous argument with $x$ and $y$ interchanged gives \begin{align*} E_M(y) \leq \|y-x\|_X+E_M(x). \end{align*} Since norms are symmetric, $\|y-x\|_X=\|x-y\|_X$, and hence \begin{align*} E_M(y) \leq \|x-y\|_X+E_M(x). \end{align*} [/step]

[step:Combine the one-sided estimates into the Lipschitz bound] From the first estimate, \begin{align*} E_M(x)-E_M(y) \leq \|x-y\|_X. \end{align*} From the second estimate, \begin{align*} E_M(y)-E_M(x) \leq \|x-y\|_X. \end{align*} Thus both $E_M(x)-E_M(y)$ and its negative are bounded above by $\|x-y\|_X$, which is equivalent to \begin{align*} |E_M(x)-E_M(y)| \leq \|x-y\|_X. \end{align*} Since $x,y \in X$ were arbitrary, $E_M$ is $1$-Lipschitz on $X$. [/step]