Let $X$ be a countably compact space such that every subspace of cardinality $\omega_1$ is metrizable. If $X$ is second countable, is $X$ metrizable? I am going through this paper:
Dow, A. An introduction to applications of elementary submodels to topology. Topology Proc. 13 (1988), https://topology.nipissingu.ca/tp/reprints/v13/tp13102.pdf
And I was stuck on the proof of
Theorem 3.1: If every subspace of cardinality $\omega_1$
of a countably compact space is metrizable, then the space
itself is metrizable.
The author first states a proposition:
Proposition 3.2 [J]. If every subspace of cardinality
at most $\omega_1$ has countable weight (weight of a topological space is the cardinality of the smallest base) then the space itself has
countable weight.
The proof starts off by assuming $X$ is not metrizable. Then the author goes on to claim "By
3.2, we know that $X$ has a subspace $Z$ with $|Z|= \omega_1$ and
$w(Z) > \omega$".
I don't quite follow how he got that conclusion. Does $X$ having weight $\omega$ make $X$ metrizable? Or does the properties that $X$ have imply that $X$ is Hausdorff? The author hasn't assumed any separation axioms for $X$.
 A: I'm pretty much sure that Alan Dow follows the conventions in Engelking, General topology. Hence a countably compact space is Hausdorff by definition.
Then just to prove the claim is easy:
W.l.o.g. $|X| > \omega$. (Otherwise, $X$ would be compact, hence regular, hence metrizable.)
Assume there is no such $Z$. Then $w(X) = \omega$ by 3.2. In particular, X is first countable. Hence $X$ is regular (see, for instance, here). Hence $X$ is metrizable.
Addendum: If $|X| > \omega$ the above also holds without assuming that countably compact spaces are Hausdorff: See here, Lemma 1.1.
On the other hand, it is obviously false for the indiscrete topology on a countable space with at least two elements.
A: Here, compact and countably compact spaces are supposed Hausdorff.
If $X$ is countably compact and not metrizable then $X$ is not second countable, i.e.  $X$ has uncountable weight (hence by Proposition 3.2, it has a subspace $Z$ like claimed by the author).
This is because a countably compact and second countable space is compact, and a compact and second countable space is metrizable.
