A subspace of $\omega_1$ is metrizable iff it is nonstationary I want to prove the following statement:

A subspace $X$ of $\omega_1$ is metrizable if and only if $X$ is nonstationary in $\omega_1$.

I haven't really achieved much:
If $X$ is stationary, then $X$ is uncountable and it is easy to see that $X$ is not compact (not even Lindelöf).
So it would suffice to show that $X$ is sequentially compact. This however need not be the case as seen by the set $X=\omega_1\setminus\{\omega\}$ and sequence $x_n=n$.
For the other direction, the only metrization theorem I'm familiar with is not applicable since if $X$ is nonstationary it need not be second-countable.
It is however clear if $X$ is bounded, since then it is a subspace of a countable ordinal, which is metrizable.
 A: A subset of $\omega_1$ is assumed to be topologized as a subspace of $\omega_1$ with the order topology.
Lemma 1. If $X$ is a meta-Lindelöf subspace of $\omega_1,$ then $X$ is nonstationary.
Proof. Since $X$ is meta-Lindelöf, the collection of all bounded open subsets of $\omega_1$ can be refined to a point-countable open cover $\mathcal G$ of $X.$ For each $\xi\in X$ choose $G_\xi\in\mathcal G$ with $\xi\in G_\xi.$ If $X$ were stationary, then $X'=\{\xi\in\omega_1:\xi\text{ is a limit point of }X\}$ would be a closed unbounded set, and $X\cap X'$ would be stationary; so it will suffice to show that $X\cap X'$ is nonstationary, by defining a divergent regressive function $f$ on $X\cap X'.$
For each $\xi\in X\cap X'$ choose $f(\xi)\in G_\xi\cap X,\ f(\xi)\lt\xi.$ I claim that $f$ is divergent. Consider any $\alpha\lt\omega_1;$ we have to show that the set $Z=\{\xi\in X\cap X':f(\xi)\lt\alpha\}$ is bounded. Now $Z\subseteq\bigcup\{G\in\mathcal G:G\cap X\cap\alpha\ne\emptyset\}.$ Since $\mathcal G$ is point-countable, we have $|\{G\in\mathcal G:G\cap X\cap\alpha\ne\emptyset\}|\le|X\cap\alpha|\cdot\aleph_0\le\aleph_0.$ So $Z$ is covered by countably many bounded sets, and is therefore bounded.
Lemma 2. If $C$ is a closed unbounded subset of $\omega_1,$ then $\omega_1\setminus C$ is metrizable.
Proof. $\omega_1\setminus C$ is the union of pairwise disjoint countable open sets. Since $\omega_1$ is first-countable and $\text{T}_3,$ every countable subspace of $\omega_1$ is second-countable and $\text{T}_3,$ hence metrizable. Thus $\omega_1\setminus C$ is a topological sum of metrizable spaces, and therefore is itself metrizable.
Theorem. For $X\subseteq\omega_1$ the following statements are equivalent:
(1) $X$ is metrizable;
(2) $X$ is paracompact;
(3) $X$ is meta-Lindelöf;
(4) $X$ is nonstationary.
Proof. (1)$\implies$(2) is A. H. Stone's theorem; (2)$\implies$(3) is trivial; (3)$\implies$(4) by Lemma 1; and
(4)$\implies$(1) by Lemma 2.
