Suppose that $f:\mathbb R_+\to\mathbb R_+$ is Lipschitz continuous and $x,y,z$ are elements of a metric space $(X,d)$. Fix $c,k>0$.
I'm reading a paper that says: if $z\in B(x,k)\cap B(y,k)$ and $d(x,y)\leq c$ with $B(x,k):=\{x'\in X:d(x',x)\leq k\}$, then $\lvert f(d(x,z))-f(d(y,z))\rvert\leq Mc,$ for some $M>0$.
Why? Can you argue?
My attempt
Fristly, it seems that the fact that $z\in B(x,k)\cap B(y,k)$ is irrelevant.
Because $f$ is Lipschitzian, for some $M>0$, $$\lvert f(d(x,z))-f(d(y,z))\rvert\leq M \lvert d(x,z)-d(y,z)\rvert.$$
Also, it holds from the triangle inequality that
\begin{align} d(x,z)\leq d(x,y)+d(y,z)\\ d(y,z)\leq d(y,x)+d(x,z). \end{align} Hence \begin{align} d(x,z)-d(y,z)\leq d(x,y)\\ -d(y,x)\leq d(x,z)-d(y,z). \end{align}
This is equivalent to $\lvert d(x,z)-d(y,z)\rvert\leq d(x,y)$, and by hypothesis, $d(x,y)\leq c$.
Is this correct? Is there any other argument that is more efficient than this?
Thanks in advance.