Say you start with the standard construction: take closed interval [0, 1], then remove open middle third, then remove two open middle thirds from two remaining segments, then remove 4 open middle thirds from 4 remaining segments, etc countably many times (more formally, intersect all of the sets you obtain in each of the step).
Intuitively, what remains should be the end points of the removed intervals, but this is not true: since at every step you removed finitely many intervals, and there were countably many steps, you removed finitecountable <= countablecountable = countably many intervals, so there are only countably many many ends of removed intervals, but the Cantor's set is uncountable. So where do the remaining points come from?
It could be that the author was thinking about sequential compactness: every sequence of elements of the space has a convergent subsequence (with its limit also in the same space).
For metric spaces, sequential and usual compactness coincide:
That's of course also true of a compact space, the author was being imprecise there. What they meant was definitely sequential compactness, as mentioned by a sibling comment.
https://archive.org/details/counterexamplesi0000gelb