Let me try to answer 3. first. It is not even clear that we can state the result you have in mind. Indeed how do you define “finite iteration“ without the finite ordinals under hand ?
And if you already have the finite ordinals, then the question becomes trivially “yes“ because to construct a given finite ordinal as a finite iteration, just use that ordinal as an indexation. However this does not reflect (or so I think) the spirit of your question, which seems to be asking about ”honest” finite iteration, like $1,2, 3$ “and so on“. However the whole problem lies in this “and so on”, which we cannot make precise without appealing to finite ordinals, and then we enter a loop.
Here's one way to answer negatively. Assuming ZF(C) is consistent, it has a model $M$ such that externally, the set of finite ordinals contains a subset on which the membership relation has the order type of $\mathbb Q$. In particular this means that the “finite ordinals“ of this subset cannot be attained by an ”honest finite iteration“. But then, you might say that this model $M$ is artifical, and not the “real“ one : I would agree, but how do you know that from inside of $M$ ?
So the answer to 3. is essentially that you will not be able to prove that in any nontrivial meaningful way.
But all hope is not lost. You can still perform induction on those finite ordinals, and prove that they all belong to $I$, where $I$ is any inductive set (which will allow you to conclude for 4. So your reasoning can be saved !
If you know about transfinite induction, then it should be clear what to do : prove by induction on all ordinals $\alpha$ the formula “$\alpha\in I$ or $\alpha$ is not finite“ which should not be too complicated.
If you do not know about transfinite induction, it's not a problem either, you only need to know that the class of ordinals is itself well-ordered. Once you have that, you can simply ponder : if $I$ is an inductive set, what is the least ordinal that doesn't belong to it ? (If there is any ! - but if there isn't well certainly all finite ordinals belong to it; although if you keep learning about ordinals you'll see that this situation doesn't happen)
Best Answer
There are various ways to show this:
Probably the most natural is to apply the replacement axiom (scheme): considering the formula "$x$ is the predecessor of $y$" we see that if the class of successor ordinals were a set, then the class of all ordinals would also be a set.
We could "close downwards" via the axiom of union: can you show that if $S$ is a set of ordinals then $\bigcup S$ is the set of all ordinals smaller than some element of $S$? Now think about the downwards closure of the class of successor ordinals.
We could consider the class of successor ordinals itself as a well-ordering with respect to "$\in$" - if it were a set it would have to be in bijection with some ordinal, but at the same time we can show that the class of successor ordinals is order-isomorphic to the class of ordinals (consider $\alpha+1\mapsto\alpha$).
In the $\mathsf{ZF}$-style context, we can just observe that for every ordinal $\alpha$ there is a successor ordinal not in $V_\alpha$ - and so the class of successor ordinals isn't a set since every set lies in some $V_\alpha$.