The ordinals are still well-ordered even without the axiom of choice, and they are still well-founded. This means that as a topological space $\alpha+1$ is still compact, and if $\alpha$ is a limit ordinal then $\alpha$ is still not compact.
Sequential compactness talks about countable subsets, so if we take $\omega_1$ it is closed under countable limits and therefore sequentially compact, but as a limit ordinal it is not compact. Note that for that to be true we need to assume a tiny bit of choice - namely $\omega_1$ is not a countable union of countable ordinals.
Without the axiom of choice we can have strange and interesting counterexamples, though. One of them being an infinite Dedekind-finite set of real numbers. Such set cannot be closed in the real numbers so it cannot be compact. However every sequence has a convergent subsequence because every sequence has only finitely many distinct elements.
There is a section in Herrlich's The Axiom of Choice in which he discusses how compactness behaves without the axiom of choice. One interesting example is that in ZFC compactness is equivalent to ultrafilter compactness, that is every ultrafilter converges.
However consider a model in which every ultrafilter over $\mathbb N$ is principal. In such model the natural numbers with the discrete topology are ultrafilter compact since every ultrafilter contains a singleton. However it is clear that the singletons form an open cover with no finite subcover.