I would like to prove the set of all countable ordinals is actually a set.
There are some related questions, but I think none of them answer exactly what I want. There is one in particular here which tries to show the result (probably it proves it) but I don't understand.
Following the above answer, I pick a set $A$ such that $|A|=\aleph_0$. The family of all well-orders of $A$ is a subclass of $A\times A$ and thus is a set. Call it $\mathscr R$. For each $R\in\mathscr R$ the well-ordered structure $(A,R)$ is order-ismorphic to some ordinal $\alpha$ (which must be countable). This proves all ordinals which are order-isomorphic to some structure $(A,R)$ is a countable set.
However, I'm not sure such a set ($\mathcal A$) is $\{\alpha\in\mathcal O : |\alpha|\leq \aleph_0\}$ whole, but only a subset. In fact, I'm very sure about that, because $\mathscr R$ is countable and: hence $\mathcal A$ and also $\bigcup\mathcal A$ will be, while the last should be $\omega_1$.
What I'm following this pourpose? Because I would like to define
$$ \omega_1 = \bigcup \{\alpha\in\mathcal O : |\alpha|\leq \aleph_0\}, $$ whithout using $\aleph_1$. It will be an ordinal because the union of a set of ordinal it is. However, it may be $\mathcal O$, the class of all ordinal numbers. The unique way to avoid this is proving $\{\alpha\in\mathcal O : |\alpha|\leq \aleph_0\}$ is a set.
Correction: As @NoahSchweber says, each $R\in \mathscr R$ is countable but $\mathscr R$ itself is a subset of $P(A\times A)$, so it may be uncountable. However, I'm not worried about countability. Showing the set is uncountable is quite easy, since if not it should be an element of itself.
I want to prove $\{\alpha\in\mathcal O : |\alpha|\leq \aleph_0\}$ is a set.