Yes, the converse is true, but only for a somewhat ridiculous reason: "finite" means "in bijection with an element of $\omega$".
However, it is possible to construct (assuming $ZFC$ is consistent) a model of $ZFC$ in which $\omega$ has an element that is intuitively infinite; this model would be ill-founded, because the immediate predecessors of that element would form a decreasing $\in$-chain, but the key is that the set of those immediate predecessors is not in the model. So as far as the model is concerned, the Axiom of Regularity is satisfied.
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
Instead of working "backwards", work "forward". In that case, the need for the axiom of infinity is also mitigated.
Begin by defining ordinals, or cardinals whichever you like. Ordinals work better since they have an inherent successor operation. Now define the notion of "finiteness", it is true that many times this notion makes an appeal to the natural numbers but there are purely set theoretical definitions of finiteness. For example Dedekind-finiteness, Tarski-finiteness, and others.
Now define $\omega$ as the collection of all finite ordinals. This is exactly what you get when you begin with $\{0\}$ and close it under the successor operation. Next show, if you like, that this set is inductive and that it is in fact minimal.
In the course I'm TA'ing right now, the professor (who is a distinguished set theorist) is taking this sort of approach. The natural numbers are some atomic object at first, and after defining the ordinals we take the finite ordinals, $\omega$ as the set theoretic natural numbers. This circumvents the need to talk about inductive sets at all.