non-expanding map and the diameter of subset of the image and complement of the image

Question

Consider a metric space $M=(X,d)$ and a map $f:X\to X$ such that for all $x,y\in X$, $$ d(f(x),f(y))\leq d(x,y). $$

Is the following statement true (maybe for finite $M$ or even compact $M$?)?

For every $Y\subseteq X$, we have $$ \text{diam}((X\setminus f(X))\cup f(Y))\geq \text{diam}(Y). $$

In particular, I'm looking for the case when the $M$ is a finite tree (simply connected finite simplicial 1-complex).

Answer 1

This is false for $M={\mathbb N}$ (with the standard metric) and $S=\{1,2\}$, $f(S)=\{1\}$, $f(M)=M$, where $f$ is a translation when restricted to $[2,\infty)$. Do not know about the compact case.

Answer 2

Nian Si showed a counterexample which I will put it here.

Consider $3$ points $\{0,1,2\}\subset \mathbb{R}$ which forms a metric space.

Consider the mapping $f(0)=1, f(1)=0$ and $f(2)=1$, and $Y=\{0,2\}$. $\text{diam}(Y)=2$.

$X\setminus f(X) = \{2\}$, $f(Y)=\{1\}$. So $\text{diam}((X\setminus f(X))\cup f(Y))=1$.