Prove $\{\alpha\in\mathcal O : |\alpha|\leq \aleph_0\}$ is a set 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.
 A: Your claim that $\mathcal{R}$ - the set of well-orderings with domain a subset$^*$ of some fixed countably infinite set $A$ - is countable is incorrect. Indeed, we can see directly that every countable ordinal is isomorphic to some element of $\mathcal{R}$:


*

*Fix $\alpha$ a countable ordinal.

*Since $\alpha$ is countable, we get an injection $f:\alpha\rightarrow A$. Let $I$ be the image of $f$.

*Define a relation $\trianglelefteq$ on $I$ as follows: $$x\trianglelefteq y\iff f^{-1}(x)<f^{-1}(y)$$ (where "$<$" is the usual ordering on $\alpha$).

*$\mathcal{L}=(I,\trianglelefteq)$ is a well-ordering with domain $\subseteq A$, so $\mathcal{L}\in\mathcal{R}$; and clearly $\mathcal{L}$ and $\alpha$ are isomorphic (the original map $f$ is an isomorphism).
From here we immediately get that the set of countable ordinals exists (or if you prefer, that the class of countable ordinals is a set), by replacement: 


*

*It's a standard theorem (using replacement) that every well-ordering is isomorphic to an ordinal.

*OK, now apply replacement to $\mathcal{R}$ to get the set of ordinals isomorphic to some element of $\mathcal{R}$; this is exactly the set of countable ordinals.

$^*$Why "a subset?" Well, this is because otherwise we miss the finite ordinals (clearly no finite ordinal is isomorphic to a well-ordering with infinite domain!). It's not a huge issue, but let's phrase things this way.
A: A binary relation on $S$ is a subset of $S\times S.$ A well-order $<$ on $S$ is a certain kind of binary relation on $S$, although we prefer to write $a<b$ rather than $(a,b)\in <.$ 
Let $A$ be countably infinite. Define $B$ by:  $b\in B$ iff  there exists $S\subset A$ such that $b$ is a well-order on $S$. We have: 
(1). If $b\in B$ there exists a $unique$ countable ordinal $x$ which is order-isomorphic to $b.$ 
(2). If $x$ is a countable ordinal there exists $b\in B$ which is order-isomorphic to $x.$
By (1),  Separation and Comprehension imply there exists the  set $C$ of all countable ordinals that are order-isomorphic to members of $B$. By (2), every countable ordinal belongs to $C.$ So $C$ is the set of countable ordinals.  
