Non-constructive axiom of infinity

Let $A$ be set with $A\ne\emptyset$ and $A=\bigcup A$. We can define the notion of ordinal, i.e. transitive well-ordered (by $\in$) sets. Even without the Axiom of Infinity, the class of ordinals is proper and also well-ordered, hence there exists a smallest ordinal $\alpha$ that does not allow an injective map $\alpha\to A$. Then $\alpha$ cannot be finite, hence infinite ordinals exist. Hence there exists a smallest infinite ordinal $\omega$.

I claim that is suffices to show that such set is infinite. We can prove that there are only finitely many sets of a given finite rank, and they are all finite. So an infinite set must have an infinite rank, so $\omega$ must exist, which is an inductive set.

Let us that if $x$ a non-empty set of a finite rank then $\bigcup x\neq x$. $\DeclareMathOperator{rank}{rank}$

It suffices to show that $\rank(\bigcup x)<\rank(x)$, when $x$ is non-empty and has a finite rank.

Recall that for finite sets it holds that $\rank(x)=\max\{\rank(y)+1\mid y\in x\}$. Therefore if $a\in\bigcup x$ then there is some $b\in x$ such that $a\in b$, so $\rank(a)+1<\rank(x)$.

Therefore $\max\{\rank(a)+1\mid a\in\bigcup x\}=\rank(\bigcup x)<\rank x$.