Proof that $\omega_1$ is countable (where is the flaw?) I have discovered a "proof" of the fact, that $\omega_1$ can be countable, contradicting its definition. It mustn't assume axiom of choice, so I guess there are some parts of this construction which can't be made without it. Anyway, here is my argument:
It's relatively consistent with ZF that $\omega_1$ is countable union of countable ordinals (here you can find this claim). Let's say that $\omega_1=\bigcup\alpha_n$. For every countable ordinal $\alpha$ we can find subset $S_\alpha$ of real numbers with order type $\alpha$. Let's construct sets $S_{\alpha_n}$, and without loss of generality assume that $S_{\alpha_n}\subset(n,n+1)$. Now, if we take union of all sets $S_{\alpha_n}$ we get a subset of real numbers with order type $\omega_1$ (from our initial assumption on $\alpha_n$). But we can show that every well ordered subset of reals is countable: let $r_\alpha$ be $\alpha$-th number in this set. Rationals are dense in real numbers, so, as $r_\alpha<r_{\alpha+1}$, we can find rational $q_\alpha$ with $r_\alpha<q_\alpha<r_{\alpha+1}$. From this it follows that we can inject $\omega_1$ on $\mathbb{Q}$ by using map $\alpha\rightarrow q_\alpha$, proving $\omega_1$ subset of countable set, thus countable.
Where is the flaw in that argument?

The proof of "Union of countable set is countable" requires the axiom of countable choice, but "$\omega_1$ is countable union of countable sets" is relatively consistent with only ZF, not ZF + countable choice.

Note that $\omega_1$ is not embeddable even if you does not assume the choice. The set $\bigcup_n S_{\alpha_n}$ is not order-isomorphic to $\omega_1$. Especially, there is no reason that the cardinality of $\bigcup_n S_{\alpha_n}$ is equal to $\aleph_1$.
