The basic Fraenkel model

Let $B\subseteq A$ in the model and let finite $E$ support it. If two atoms $a,b\mathrm{\notin }E$ had different membership in $B$, their transposition would fix $E$ but move $B$. Thus either no atom outside $E$ lies in $B$, making $B$ finite, or every atom outside $E$ lies in $B$, making it cofinite.

The set $A$ is infinite because every ground finite subset omits an atom. If $f:\omega \to A$ were injective in the model, the even-indexed range would be infinite and its complement would contain the infinite odd-indexed range, contradicting step 1.1. Hence there is no such injection, in agreement with F2.

If a well-order $<$ of $A$ had finite support $E$, the nonempty invariant set $A\setminus E$ would have a least member $a$. Choose $b\mathrm{\notin }E\cup \{a\}$ and transpose $a,b$. The transposition fixes $<$ and $E$, so must send its uniquely least outside-$E$ member to itself, but sends $a$ to $b$, contradiction. Thus $A$ is not well-orderable. F3 now shows by contraposition that AC fails in the permutation model. This reductive use of AC is the only choice dependence of the conclusion.