= Solution
A <sequentially compact space> is one in which every sequence has a subsequence converging to a point of the space. A <limit point> $\ell$ of $S$ means every open ball around $\ell$ meets $S\setminus\{\ell\}$.
In a compact <metric space>, a sequence with a finite set of values has a constant subsequence. If its value set is infinite and had no <limit point>, each point of the space would have a neighbourhood containing at most one point of that value set. A finite subcover would then make the value set finite, a contradiction. It therefore has a <limit point> in the space, and the permitted subsequence result gives a convergent subsequence. This proves sequential compactness.
For the remaining assertions assume the <compact space> is nonempty, as required for a fixed-point claim. The displacement function $q(x)=d(f(x),x)$ is continuous by the stated continuity of the metric and of $f$. The <extreme value theorem> gives a minimum $\delta=q(p)$. If $\delta>0$, then $p\neq f(p)$, and strict decrease of distances implies
$$
q(f(p))=d(f^2(p),f(p))<d(f(p),p)=\delta,
$$
contradicting minimality. Thus $\delta=0$ and $p$ is a <fixed point>. Two distinct fixed points would satisfy $d(p,q)=d(f(p),f(q))<d(p,q)$, also impossible. Hence the <fixed point> is unique.
On the noncompact real line, take
$$
\boxed{f(x)=\log(1+e^x).}
$$
Its derivative $e^x/(1+e^x)$ is strictly between zero and one. The <mean value theorem> therefore gives $|f(x)-f(y)|<|x-y|$ for each distinct pair. Yet $f(x)-x=\log(1+e^{-x})>0$, so it has no <fixed point>. Compactness, rather than merely pointwise strict shrinkage, supplies attainment of a zero displacement in the preceding proof.
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