Topological properties of $[0,\omega_1)$ without choice. Reading this wikipedia article, I arrived at the fact that $\omega_1$ can exist without choice. Since the proof I know of the fact that $[0,\omega_1)$ is sequentially compact depends on the fact that the countable union of countable sets is countable, I arrived at the following question:

What topological properties of $[0,\omega_1)$ depend on the axiom of choice?

 A: First of all, yes. $\omega_1$ exists without assuming the axiom of choice. However it can be the countable union of countable sets, which can cause some troubles. Let me just point out that $\omega_1$ is, by definition, the set $[0,\omega_1)$. So I'll just use the shorter notation to make it easier.
So if $\omega_1$ is the countable union of countable sets, then it is the limit of an $\omega$-sequence of countable ordinals. Therefore it will no longer be the case that every sequence of countable ordinals has a countable limit. Therefore it will not be sequentially compact.
To see why this is true, if $\omega_1=\bigcup A_n$, where each $A_n$ is countable, there is a unique ordinal isomorphic [uniquely] to $A_n$. Now let $\alpha_n$ be that ordinal, and let $\beta_n=\alpha_1+\ldots+\alpha_k$. Then $\beta_n$ are increasing, and $\sup\beta_n=\lim\beta_n=\omega_1$.
For metrizability, the answer is that the axiom of choice is not needed to prove that $\omega_1$ is not metrizable. This is a combination of Corollaries 4.2 and 4.4 in the following paper:

C. Good and I. J. Tree, Continuing horrors of topology without choice, Topology Appl. 63 (1995), no. 1, 79--90.

