contractive maps are uniformly continuous

Theorem A contraction mapping is uniformly continuous.

Proof Let $T:X\to X$ be a contraction mapping in a metric space$X$ with metric $d$. Thus, for some $q\in [0,1)$, we havefor all $x,y\in X$,

$$d(Tx,Ty)\le qd(x,y).$$

To prove that $T$ is uniformly continuous, let $\epsilon >0$ be given.There are two cases.If $q=0$, our claim is trivial, since then for all $x,y\in X$,

On the other hand, suppose $q\in (0,1)$. Then for all $x,y\in X$ with, we have

In conclusion, $T$ is uniformly continuous. $\mathrm{\square }$

The result is stated without proof in [1], pp. 221.