Would someone please enlarge on Arturo Magidin's original answer ?
$1.$ Say the question didn't divulge $|S| = |X|$. Then how can $|S|$ be determined? Any intuition?
I recast it below with more details to try to apprehend it. Please feel free to edit errors/problems.
As defined therein, $X$ is an infinite set, and $ S := \{$ all finite subsets of $X \}$.
Since ● $X$ is given as infinite which means $|X| \ge |\mathbb{N}|$
and ● $S \quad \supseteq \quad \{\emptyset, \{1\}, \{2\},...,\{n\},...\} = \require{cancel} \cancel{\{\text{singletons}\}} = |\mathbb{N}|$
$\implies \emptyset, \{1\}, \{2\},...,\{n\},... \in S$
$\implies 1, 2, ..., n, ... \in X \iff \mathbb{N} \subseteq X$ thus $\color{limegreen}{|\mathbb{N}| \le |X| \le |S|}$.
For all $n \in \mathbb{N}$, define $S_n$ as did Prof Magidin: $S_n:= \color{violet}{\{}{\text{ all subsets of cardinality $n$ }} \color{violet}{\}}\subseteq S.$
This definition involves a subset of cardinality $n$ (ie with $n$ elements) so we now scrutinise it.
$2.$ How would you divine/previse to define $S$ wrt $S_n$ ?
$S_n$ has the form $\color{violet}{\{} \quad \{ \, (a_1, ..., a_n) \, \}, \{ \, (b_1, ..., b_n) \, \}, ...\quad \color{violet}{\}}$.
The $n$ elements in each $n$-tuple in each subset of cardinality $n$ can be chosen $n!$ ways,
ie: the $a_1, ..., a_n$ in $(a_1, ..., a_n)$ in $\{ \; (a_1, ..., a_n) \; \}$ can be chosen $n!$ ways.
Thus every such subset of cardinality $n$ determines $n!$ $n$-tuples of elements of $X$.
In general, an $n$-tuple included in any set will produce a subset (of that set) of cardinality $\le n$.
This paragraph implies: $ |S_n| \color{red}{\le} \, n!|X|^n \, = |X|$.
$3.$ How and why is $ |S_n| \color{red}{\le} \, n!|X|^n $?
In toto, $ \begin{align} |X| \color{lime}{\le} |S| & = \left|\biguplus_{n=0}^{\infty} S_n\right| = \sum_{n=0}^{\infty}|S_n| = |\underbrace{S_0}_{= \emptyset}| + \sum_{n=1}^{\infty}|S_n| \\ & = \sum_{n=1}^{\infty}|S_n| \\ & \color{red}{\le} \sum_{n=1}^{\infty}|X| = |\mathbb{N}||X| \; {\Large{\color{darkturquoise}{[}}} = {\Large{{\color{darkturquoise}{]}}}} \; |X| \end{align} $
$4.$ In the last line, how and why is $|\mathbb{N}||X| \; = \; |X|$ ?