Proof
(related to Proposition: Subsets of Finite Sets)
 If \(S\subseteq X\) and \(S=X\) then \(S\) is finite, since \(X\) is finite by hypothesis.
 Let \(S\subset X\) and \(S\neq X\).
 Since \(X\) is finite by hypothesis, there is an \(n\in\mathbb N\) with \(X=n\).
 Since \(S\subset X\), \(S < X\), by characterization of finite sets $(3)$ and comparison of cardinal numbers.
 Let \(K\) be the set of all natural numbers, with \(S \le k, k\in K\). Since \(n\in K\), the set \(K\) is nonempty.
 Following the wellordering principle, \(K\) has a minimal element \(k_0\in K\), for which \(S \le k_0\).
 Note, that \(S \not < k_0\). Otherwise for some \(j < k_0, j\in K\) we would have \(S \le j\), contradicting the minimality of \(k_0\).
 Therefore \(S = k_0=k_0\).
 Thus, \(S\) is finite.
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References
Bibliography
 Reinhardt F., Soeder H.: "dtvAtlas zur Mathematik", Deutsche Taschenbuch Verlag, 1994, 10th Edition