 Solution

A contraction satisfies $d(Tx,Ty)\le qd(x,y)$ for some $0\le q<1$. Starting from $x_0$, define $x_{k+1}=T{x}_k$. The geometric bound on successive distances makes $(x_k)$ Cauchy, so completeness gives a limit $x$. Continuity of $T$ gives $Tx=x$. If $Ty=y$, then $d(x,y)\le qd(x,y)$, hence $x=y$. This is the contraction mapping theorem.

If $T^n$ is a contraction, it has a unique fixed point $x$. Since $T^n(Tx)=T(T^{n}x)=Tx$, the point $Tx$ is also fixed by $T^n$, so uniqueness gives $Tx=x$. Every fixed point of $T$ is fixed by $T^n$, so it too must equal $x$. Thus $T$ has exactly one fixed point.