The space $X$ is sequential if for each nonclosed $A \subset X$, there exists a convergent sequence $a_n \rightarrow x$ so that $a_n \in A$ but $x \notin A$. By $\omega_1$ we mean an uncountable well ordered set (with the order topology) so that each bounded subset is countable. Sequentially compact means each sequence in $X$ has a convergent subsequence.
Suppose $X$ is sequential, sequentially compact, and Hausdorff (or weakly Hausdorff). If $X$ is not compact, must $X$ contain a subspace homeomorphic to $\omega_1$?
(The converse is true. If $X$ is compact then $\omega_1$ cannot embed in $X$.)
There is a well studied space of Ostaszewski which is plausibly? obviously? obviously not? a counterexample:
A. J. Ostaszewski, “On countably compact, perfectly normal spaces,” Journal of the London Mathematical Society, volume 14, (1976), pp. 505–516
It is apparent that attempting to employ sequential compactness in an effort to characterize the compactness property of sequential spaces can take us outside ZFC.
[Vaughan(1984)] J. E. Vaughan, “Countably compact and sequ entially compact spaces,” inHand-book of Set-Theoretic Topology (edited by K. Kunen and J. E. Vaughan), pp. 569–602
Corollary 6.12 of Anthony Goreham's arxived paper http://arxiv.org/pdf/math/0412558v1.pdf, asserts: The statement ‘Compactness and sequential compactness are equivalent for perfectly regular sequential spaces’ is independent of ZFC.