Let $(X,\leq)$ be a totally ordered set such that, equipped with the order topology, $X$ is compact. Is then $X$ homeomorphic to a closed subset $A \subseteq [0,1]$?
A way to ask this question colloquially: Can we put points together in a compact way such that they are "denser" than the reals? I feel the answer is no but I lack a proper counterexample.
No: the closed long ray is a counterexample. It is a totally ordered set with the order topology, it is compact, and it is not metrizable. Another, perhaps easier, counterexample is the lexicographically ordered square, i.e., $[0,1]\times[0,1]$ with the lexicographic order topology. This is compact, but
$$\big\{\{x\}\times(0,1):x\in[0,1]\big\}$$
is an uncountable, pairwise disjoint family of open sets, something that cannot exist in the separable space $[0,1]$.