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!

The union of any proper class of ordinals is the class of all ordinals, actually, which is not, itself, an ordinal. If this were not so, then there would be a least ordinal $\alpha$ not in the union. Then $\alpha$ would necessarily be an upper bound on the ordinals in the class, so the class is contained in the set $\alpha\cup\{\alpha\}$, and so not proper, at all.
For your first question, consider the set $$\mathcal{W}=\bigl\{\langle X,R\rangle\in\mathcal{P}(\omega)\times\mathcal{P}(\omega\times\omega):R\text{ well-orders }X\bigr\}.$$ Every countable well-order type is represented by at least one of the members of this set, so by replacement, the collection of countable well-order types (ordinals) is a set.
