I'm trying to read this proof from Dow's An Introduction to Applications of Elementary Submodels to Topology. (Proposition 3.1: If $X$ is countably compact and every subspace of cardinality at most $\aleph_1$ is metrizable then so is $X$.)
I haven't been able to verify a claim in the fourth paragraph. Namely that for each $\alpha\in\omega_1$ we have $\exists x\in \overline{X\cap M_\alpha}$ such that $\tau\cap M_\alpha$ does not contain a base for $x$. (It's curious to me here that "contain a base" and "is a base" seems to be equivalent, so I don't get why word it as "contains a base".)
I think this is all the context that is necessary:
$(M_\alpha)_{\alpha\in\omega_1}$ is a continous $\in$-chain of countable elementary submodels of the universe with $(X,\tau)\in M_0$.
$M$ is their union (so it has size $\aleph_1$ and is $\omega$-covering).
$X$ is countably compact, and we have supposed by contradiction every subspace of $X$ of cardinality at most $\aleph_1$ is metrizable but $X$ is not. In particular $X\cap M$ is metrizable.
I think I can show that if $\tau\cap M_\alpha$ is a base for $\overline{X\cap M_\alpha}$ then $X=\overline{X\cap M_\alpha}$, hence the title of this question. ($X$ second countable would imply metrizable as it is countably compact.)
Short version: Can you help me verify/contradict the statement in the title? (:
The question in the title must be answered negatively: the Sorgenfrey line has $\mathbb{Q}$ as a countable dense metrisable subset (it is homeomorphic to $\mathbb{Q}$ in the standard topology as it is metrisable by Urysohn and has no isolated points, e.g. see this survey paper) but is not metrisable as it has weight $\mathfrak{c}$.