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There are old ZFC examples due to Eric van Douwen that satisfy all the properties in the title, except that they are of cardinality $2^{\aleph_0}$, so the answer to the title question is YES if the Continuum Hypothesis (CH) is assumed, but I am asking for an example which does not use any axiom beyond the ZFC axioms.

It would be especially nice if it were first countable, like Eric's examples, and normal, like one of his examples. A Souslin tree with the interval topology qualifies, but the existence of Souslin trees is ZFC-independent. I have been able to weaken CH to "stick" [which says that there is a family of $\aleph_1$ countable subsets of an uncountable set, such that every uncountable subset contains a member of the family] and I also have an example if $\mathfrak b = \aleph_1$ but no ZFC example.

It would also be interesting to know whether the existence of an example implies the existence of a first countable example. There are lots of first countable examples under CH, and the "stick" example I have in mind is also first countable, as is my $\mathfrak b = \aleph_1$ example. For sure, there is a scattered example if there is one at all: the "Kunen line" qualifies under CH, while if CH is negated, then we use the fact that every crowded (= dense-in-itself) compact Hausdorff space is of cardinality at least $2^{\aleph_0}$.

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    $\begingroup$ By the way, "$\omega_1$-compact" means every closed discrete subspace is countable, while $\sigma$-countably compact means it is the union of countably many countably compact subsets. $\endgroup$
    – Peter Nyikos
    Commented Jan 13, 2016 at 0:10
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    $\begingroup$ Peter Nyikos. Welcome to mathoverflow. $\endgroup$
    – Joseph Van Name
    Commented Jan 13, 2016 at 1:13
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    $\begingroup$ Hello Peter. Just to check if I still remember some things about this (so this comment does not help to settle the question). The axiom named "CC" that you considered in a 2009 article with T. Eisworth implies that if such a space exists, then there exists a separable one. Am I right ? $\endgroup$
    – Mathieu Baillif
    Commented Jan 13, 2016 at 23:14
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    $\begingroup$ Thanks for the welcome, Joseph. Perhaps you are right, Mathieu, but I don't see it offhand. The first trichotomy axiom there is a consequence of CC and says, in effect, that every such space has a countable subset with non-Lindel\"of closure. The closure must still be $\omega_1$-compact but I see no reason why it cannot be countably compact and noncompact, hence non-Lindel\"of. Can you? $\endgroup$
    – Peter Nyikos
    Commented Jan 14, 2016 at 1:08
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    $\begingroup$ No, I don't have a proof that the closure is not sigma-countably compact. It's just that I wrote my comment too late in the night yesterday and the basic fact that this property does not transfer from the whole space somehow escaped me. $\endgroup$
    – Mathieu Baillif
    Commented Jan 14, 2016 at 10:13

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Lyubomyr Zdomskyy has solved this problem. He has shown:

Theorem. If P-Ideal Dichotomy (PID) holds and $\mathfrak p > \aleph_1$, then every locally compact, $\omega_1$-compact space of cardinality $\aleph_1$ is $\sigma$-countably compact.

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