= Solution
The strict distance inequality implies $d(fx,fy)\leq d(x,y)$ for all points, so $f$ is <Lipschitz continuous>. Hence
$$
g(x)=d(fx,x)
$$
is a <continuous function>, since $|g(x)-g(y)|\leq d(fx,fy)+d(x,y)\leq2d(x,y)$. On the nonempty <compact metric space>, $g$ attains its minimum at some $p$.
If $fp\ne p$, the strict distance inequality applied to $fp,p$ gives
$$
g(fp)=d(f^2p,fp)<d(fp,p)=g(p),
$$
a contradiction. Thus $fp=p$. If $p,q$ were different <fixed points>, then $d(p,q)=d(fp,fq)<d(p,q)$, again impossible. \b[There is exactly one fixed point.] This proves <strict distance decrease on a compact metric space> without assuming a uniform contraction constant.
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