Q:

Is $\omega_1$ metrizable?

Following Urysohn's metrization theorem, I would like to prove or disprove that $\omega_1$ is metrizable. I know it is hausdorff, but I'm not sure whether or not it is second countable, and I'm at loss on how I can prove or disprove it is second countable. And if I'll disprove it, will it tell me that $\omega_1$ is not metrizable, or is Urysohn's theorem only an 'if' and not an 'iff'? 
Or perhaps there is a more comfortable-to-use in here theorem to prove/disprove $\omega_1$ being metrizable?

A:

(1) $\omega_1$ is not only Hausdorff, it is normal, as is every linearly ordered set in its order topology.
(2) $\omega_1$ is not second countable, because the intervals $[0,\alpha)$ for $\alpha\lt\omega_1$ are an open cover with no countable subcover.
(3) $\omega_1$ is not separable. Proof: Let S be any countable subset of $\omega_1$. Then $\sup S\lt\omega_1$. Let $\alpha=\sup S$. Then $S$ is a subset of the closed set $[0,\alpha]$, so the closure of $S$ is a subset of $[0,\alpha]$ which is a proper subset of $\omega_1$, so $S$ is not dense in $\omega_1$.
(4) $\omega_1$ is sequentially compact. Proof: Every sequence in $\omega_1$ has a monotonic sequence, which converges.
(5) If a metric space is not separable, then it is not sequentially compact. Proof: Let $X$ be a metric space. If $X$ is not separable, then for some $\varepsilon\gt0$ there is an uncountable set of points such that the distance between any two is at least $\varepsilon$. An infinite sequence of those points can not converge, and it can not have a convergent subsequence. Therefore $X$ is not sequentially compact.
From (3), (4), and (5) it follows that $\omega_1$ is not metrizable. Namely, if $\omega_1$ were metrizable, then, since by (3) it is not separable, it would follow by (5) that it is not sequentially compact, contradicting (4).

