 Solution

A metric space is sequentially compact if every sequence has a convergent subsequence whose limit belongs to the space. A limit point $\ell$ of $S$ is a point for which every open ball about $\ell$ contains a point of $S$ other than $\ell$.

Let $(x_n)$ lie in a compact metric space. If one value occurs infinitely often, take a constant subsequence. Otherwise its set of values $S$ is infinite. If $S$ had no limit point, every point of the space would have a neighborhood containing at most one member of $S$. A finite subcover of these neighborhoods would force $S$ finite, a contradiction. Thus $S$ has a limit point; the stated subsequence result supplies a convergent subsequence. This proves sequential compactness.

Assume now that $X$ is nonempty, as is necessary for the requested minimum and fixed point. The function $r(x)=d(f(x),x)$ is continuous, as a composition of $x\mapsto (f(x),x)$ with the metric. The extreme value theorem on a compact space implies that $r$ attains its infimum, say at $p$. If $f(p)\ne p$, strict decrease of distances givescontradicting minimality. Thus $f(p)=p$. If $p,q$ were distinct fixed points, then $d(p,q)=d(f(p),f(q))<d(p,q)$, another contradiction. Hence there is exactly one fixed point. This is the fixed point theorem for a strictly distance-decreasing map on a compact metric space.

Compactness is essential. On $\mathbb{R}$, takeHere $f(x)-x=\log (1+e^{-x})>0$, so there is no fixed point. Nevertheless $0<f^{\prime }(x)=e^x/(1+e^x)<1$ everywhere, and the mean value theorem gives $|f(x)-f(y)|<|x-y|$ for distinct $x,y$. There is no uniform Lipschitz constant less than $1$, since $f^{\prime }(x)\to 1$ as $x\to \infty$.