Proof

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 non-empty.
Following the well-ordering 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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Reinhardt F., Soeder H.: "dtv-Atlas zur Mathematik", Deutsche Taschenbuch Verlag, 1994, 10th Edition