TITLE: Proof of: $X$ is finite $\iff X$ is Tarski-finite
QUESTION [9 upvotes]: I am self-studying Horst Herrlich, Axiom of Choice (Lecture Notes in Mathematics, Vol. 1876). In the fourth chapter, he deals with different definitions of finite set. Here is the classical one:

Definition 1. A set $X$ is called finite if there exists a bijection $n \to X$ for some $n \in \mathbb{N}$. Otherwise it is called infinite.

Here is Tarski's (1924) definition:

Definition 2. A set $X$ is called Tarski-finite, provided that each non-empty subset of $\mathcal P(X)$ contains a minimal element w.r.t. the inclusion order. Otherwise it is called Tarski-infinite.

The following holds:

Proposition. Equivalent are:

$X$ is Tarski-finite.

If $\mathcal U \subseteq \mathcal P(X)$ satisfies
a) $\emptyset \in \mathcal U$, and
b) $A \in \mathcal U$ and $x \in X$ imply $A \cup \{x\} \in \mathcal U$,


then $X \in \mathcal U$.

The proof of the proposition is simple. But right after that, he writes:

Observe that by Proposition, the definition of finiteness as given in Definition 1 is equivalent to the one in Definition 2.

No proof follows. The "Def.1 $\implies$ Def.2" part seems easy to me, but the converse doesn't. Would you help me with that? Please, note that we do not want to use AC (indeed, all this thing is about proving that the two definitions are equivalent even in ZF). Therefore, we can't even use that "$X$ infinite $\implies \aleph_0 \leq |X|$", since the latter is not a theorem of ZF.

REPLY [3 votes]: What you would like to do is to say something like "Pick some element $x_0\in X$ then by induction pick $x_{n+1}\in X_n=X\setminus\{x_0,\ldots,x_n\}$" then $\mathcal U=\{X_n\mid n\in\Bbb N\}$ is a family of sets without a minimal element.
But this is impossible, since moving from the induction to the infinite sequence requires some choice. 
But there is a guiding philosophy in many choiceless proofs: If you can't choose, take everything. So taking all the co-finite sets would work. There is no minimal co-finite set, since removing a single element from an infinite set gives an infinite set.
(I gave this exercise to my students last year, and after we talked about the axiom of choice we proved it again, in the simpler way.)