Note admins: THIS IS NOT A DUPLICATE to other posts (see the last sentence).
Prove that if $A$ and $B$ are Dedekind finite then so is their union, without using the concept of natural numbers or Axiom of Choice.
Of course, the result is trivial with the natural numbers, but the book from which this question is derived said this can be proven without the aid of natural numbers and Axiom of Choice (I am using Moschovakis' Notes on Set Theory, 1st edition, chapter 4, x4.25)
We are allowed to use $ZF$ Axioms.
Suppose on the contrary thay there are Dedekind finite sets $A,B$ such that $A \cup B$ is Dedekind infinite. If either $f(A) \subset A$ or $f(B) \subset B$ then we can easily create a proper injection from one set to itself (just take restriction of $f$), so assume that that is not the case.
I am not sure what to do here. A hint would be more appreciated then a full answer.
I realize that there already are many questions and answers regarding how a union of two Dedekind finite sets are still Dedekind finite, but my question is different: the difficulty here is that (1) since I am not allowed to use natural numbers, we cannot talk about a set being finite or infinte, much less list some elements or sets, and (2) we cannot use Axiom of Choice.