FALSE: every countably compact space is compact

False claim: every countably compact topological space (Countably compact, Lindel"of, sequentially compact, limit point compact and $\sigma$-compact spaces, and relatively compact subsets) is compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

Where the claim comes from, and what is actually true. Countable compactness tests only the at most countable open covers, and the claim above asserts that testing those is enough. It is enough when the space is also Lindelöf, and the claim above drops that hypothesis.

The refutation assumes the Axiom of Countable Choice (The Axiom of Countable Choice ($\mathrm{AC}_\omega$)), which is what makes each witness countably compact. Two witnesses are available and both are refutations on their own: the first uncountable ordinal $\omega_1$ with its order topology (The first uncountable ordinal $\omega_1 := \aleph(\omega)$, On an ordinal with its order topology the sets $[0,\beta]$ and $(\alpha,\beta]$ form a basis of clopen sets, the isolated points are exactly the non-limit ordinals, and the space is Hausdorff), and the closed long ray (The closed long ray $\omega_1 \times [0,1)$ under the lexicographic order, and the long line, with the order topology).

Why the missing hypothesis is Lindelöfness and not something weaker. Countable compactness plus Lindelöfness does give compactness (Compact implies countably compact, Lindel"of and limit point compact; countably compact together with Lindel"of implies compact; and, at the cost of countable or dependent choice, sequentially compact implies countably compact, countably compact implies limit point compact, and the converse holds when every singleton is closed, claim 1(b)), so a countably compact non-compact space must fail to be Lindelöf. The long ray is checked to fail it directly (Every closed initial segment of the long ray is compact; the long ray is not compact; and, assuming countable choice, it is countably compact and not Lindel"of, claim 4), and $\omega_1$ fails it for the same reason: the cover by initial segments has no at most countable subcover, an at most countable set of countable ordinals being bounded.

The two witnesses are not the same space and neither is redundant. The ordinal $\omega_1$ is also sequentially compact, so it separates compactness from sequential compactness as well (FALSE: every sequentially compact space is compact); the long ray is a linear continuum and is connected, so it also shows that connectedness contributes nothing to compactness.