I've doubt about Dedekind-infinite sets, sets which are in bijection with a proper part, in the ZF axiomatic framework, without Axiom of Choice.
Assume a Dedekind-infinite set X exists.
Then it can be proved X contains a Dedekind-infinite N set which satisfy the Peano Axioms.
This set N can be well ordered using traditional arguments from Peano Axioms.
It can also be proved its initial chains, in this well order, I_n = { m < n }, are Dedekind-finite.
Can be proved they are also finite using the ZF definition of "finite set", without any further assumption?
Yes. The definition of finite is being equipotent with a proper initial segment of $\omega$, that is the ordinal corresponding to $\Bbb N$.
Once you have established a bijection, and in fact an order isomorphism, between your "copy of $\Bbb N$" and $\omega$, and you have by virtue of it being a model of $\sf PA_2$, then the initial segments correspond exactly to the actual finite cardinals.
The axiom of choice is needed when you want to prove that a Dedekind-finite set is finite. This is not what you're doing here. You're not arguing that these are Dedekind-finite, you're quite literally defining them to be finite.