Two questions regarding Ordinal Numbers. I'm trying to prove that there is an uncountable ordinal all which members are countable ordinal. This is fairly easy if I can state that the class of all countable ordinals is a set and then take the union on of that set (which I already know is an ordinal). However I'm not sure how to justify the fact that it is indeed a set and not a proper class.
Secondly, I was wondering if a union of a proper class of ordinals is necessarily an ordinal in the same way a union of a set of ordinals is?
Looking forward for your answers.
Thanks!

First note that every countable ordinal is in bijection with $\omega$, therefore we can view it simply as a relation on $\omega$ which well-orders it by this order type.
Therefore we may define an equivalence relation on $\mathcal P(\omega\times\omega)$ in which every two well-orders are equivalent if and only if they are isomorphic (and everything which is not a well-ordering of $\omega$ is lumped into one big equivalence class).
Now you can show that this too makes a set, it is a definable collection of $\mathcal P(\mathcal P(\omega\times\omega))$, and since every well-ordered relation is isomorphic to a unique ordinal there is a definable map from that set onto the class of countable ordinals. Therefore countable ordinals make a set.
The union of this set is again a set, by the axiom of union, and it is an ordinal by showing it is transitive and $\in$ well-orders it.
As for the second question the answer is obviously no. If the union of a proper class of ordinals would be an ordinal then it would have to be an ordinal which is a set, which has a proper class of members. The union of any proper class of ordinals, if so, is the entire class of ordinals.

Also related: How do we know an $ \aleph_1 $ exists at all?
