[Edit: I misunderstood your definition, and read as describing an injective function. The "solution" below explains why this cannot work.]
What you are suggesting is interesting. The question is whether we can assign to each limit ordinal a cofinal sequence in such a way that the $f$ you describe ends up being injective. I do not know off hand whether this is possible, but it is a nice problem (it is not clear yet whether $f$ is well-defined, i.e., whether one can arrange so that all the relevant series actually converge). Sorry for the original confusion.
The argument you are giving cannot work. This is because in your construction you are not only trying to define an injection, but in fact a strictly increasing function.
But if $f:\Omega\to{\mathbb R}$ is increasing, then for each $\alpha$ there will be rational in the interval between $f(\alpha)$ and $f(\alpha+1)$, and different values of $\alpha$ will correspond to different rationals. Of course, this is impossible as $\Omega$ is uncountable but ${\mathbb Q}$ is countable.
In fact, you will not be able to define an injection $f$ by any explicit procedure. This is because it is consistent with the axioms of set theory without choice that no such injections exist, and it is consistent with set theory plus choice that no such injection is definable.
One way to show that there are such injections is to use Zorn's lemma on the collection of injections $f$ whose domain is an ordinal. This is a partial order: $f\le g$ if $g$ extends $f$ as a function, i.e., iff $g$ has a larger domain than $f$, and the restriction of $g$ to the domain of $f$ is just $f$. Clearly a maximal element in this poset must have uncountable domain, since the reals are uncountable.
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
Assuming you are defining an inductive set as one that contains $\emptyset$ and contains $s(\alpha):=\alpha\cup\{\alpha\}$ whenever it contains $\alpha$, you can reason as follows:
By the axiom of infinity, there is some inductive set $X$. Let $$Y=\{\alpha\in X\mid \alpha \text{ is empty or is a successor ordinal.}\}.$$ Then $\emptyset\in Y$, and if $\alpha\in Y$, then $s(\alpha)$ is a successor ordinal (since $\alpha$ is an ordinal) in $X$ (since $X$ is inductive), and so is also in $Y$, so $Y$ is inductive, and by definition $Y$ consists only of $\emptyset$ and successor ordinals.
Then in conclusion, as you observed in your question, the minimal inductive set $\mathbb N$ is contained in $Y$, hence also consists only of $\emptyset$ and successor ordinals.