= Solution
The strict distance inequality also gives $d(fx,fy)\le d(x,y)$ when $x=y$, so $f$ is <Lipschitz continuous> with constant at most one, hence <continuous>. Put $g(x)=d(x,f(x))$. The <triangle inequality> gives
$$
|g(x)-g(y)|\le d(x,y)+d(fx,fy)\le2d(x,y).
$$
Thus $g$ is also <Lipschitz continuous> and <continuous>.
Assume the <compact metric space> is nonempty, as needed for existence of a point. By the <extreme value theorem>, $g$ attains its infimum at some $p$. If $h=g(p)>0$, then $p\ne f(p)$, and strict distance decrease gives
$$
g(f(p))=d(f(p),f(f(p)))<d(p,f(p))=h,
$$
contradicting minimality. Since distances are nonnegative, $h=0$ and $f(p)=p$. If two different <fixed points> $p,q$ existed, then $d(p,q)=d(f(p),f(q))<d(p,q)$, another contradiction. Therefore \b[there is exactly one fixed point]. This is <strict distance decrease on a compact metric space>; no uniform contraction factor less than one was assumed.
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