About the proof that an infinite ordinal has the same cardinality as its successor I'm aware this result (and the standard/obvious proof) is considered basic and while I've accepted and used it numerous times in the past, I'm starting to question its validity, or rather that said proof doesn't subtly require a form of the AC. (Disclaimer: it's been some time since I've looked at set theory.)
I found the same question answered here using the same method I'd used in the past. (Why are infinite cardinals limit ordinals?) My posting here is more concerned with the apparent assumptions required to use this argument.
The (apparent) issue here is that it presupposes that $\omega$ is the least infinite ordinal, i.e. a subset of any infinite ordinal $\alpha$.  Now a countable choice argument could easily prove this, but this shouldn't be necessary, one would think.
I've considered this: a simple (natural) induction argument showing that every natural number is an element of any infinite ordinal.  This would show $\omega\subseteq \alpha$, or that $\omega$ is the least infinite ordinal. Does this seem right?
Thank you 
To clarify: I'm trying to show that $\omega$ is the least infinite ordinal.  Evidently some define it this way, (there is clearly some least infinite ordinal) but the definition of $\omega$ I'm working with involves defining inductive classes and taking their intersection, which is then axiomatized as a set.  (These two definitions should be equivalent)
 A: The definition of $\omega$ is "the least infinite ordinal". So there are no hidden assumptions.
But even ignoring that, the proof I gave in the answer you linked doesn't use any "presupposed assumption". It only uses the fact that if $\alpha$ is an infinite ordinal, then for any finite ordinal $n$, $n<\alpha$.
In any case, an ordinal is by definition a well-ordered set. So we can exploit that to define a choice function on its subsets. So usually when choice is needed to prove things in general, it can be eliminated in the case of ordinals (since we already have a choice function).

You are correct, however, that induction is needed to prove that no infinite ordinal is a predecessor of a finite ordinal. For example by proving first that a subset of a finite set is finite (which is where you'd use induction), and the noting that if $\alpha\in\beta$, then $\alpha\subseteq\beta$, and so if $\beta$ is finite, $\alpha$ must be too.
A: Hopefully you already know that $\omega$ (defined as the intersection of all inductive sets) is an ordinal in the first place.
Hopefully you also know that the ordinals are totally ordered by $\in$ -- that is, for ordinals $\alpha,\beta$ we always have either $\alpha\in\beta$ or $\alpha=\beta$ or $\beta\in\alpha$.
What is an infinite ordinal? Usually "finite" is defined to mean, "equinumerous with an element of $\omega$". So if $\alpha$ is an ordinal that is not finite, then in particular it is not itself an element of $\omega$. Therefore we have either $\omega=\alpha$ or $\omega\in\alpha$. In either case we have $\omega\subseteq\alpha$.
A: (1). In the absence of Infinity, Choice, and Foundation (Regularity):
Let $On$ be the class of ordinals. Define the class $FOn$ of finite ordinals as $\{x\in On: \forall y\in x\cup \{x\}\;(y=0\lor y\ne \cup y)\}.$ 
Prove that $\forall x\in FOn \;\forall y\in x\;(y\in FOn).$  
(1-i). That is, $FOn$ is a transitive class. 
(1-ii). Any   set or definable class of ordinals is well-ordered by $\in.$
(2). Introducing Infinity. Suppose $\exists z\in On\backslash FOn.$  For any $x\in FOn$ we have $(z=x\lor z\in x\lor x\in z).$  But $z=x\in FOn$ cannot hold, and $z\in x\implies z\in FOn$ by def'n of $FOn.$ So every $x\in FOn$ belongs to $ z .$
So by Comprehension $\exists O \;\forall x\;( x\in O \iff (x\in z\land x\in FOn)).$  That is, the set $O$ is equal to $FOn.$ 
So by (I-i) and (1-ii), and by the def'n of $On,$ we have $O\in On.$ Now for any $z'\in On\backslash FOn$ we have $z'\geq O$ because $z'<O\implies z'\in O\implies z'\in FOn,$ a contradiction. So $O=FOn$ is the least infinite ordinal. Usually denoted by $\omega.$ 
And now prove that $0\in FOn$ (i.e. $FOn$ is not empty) and that $\forall x \in FOn\;(x+1\in FOn).$    
