Cardinality of the Set of all finite subset of $\mathbb{R}$

We can show that $\#S \ge {c}$ via the injection $f: \mathbb{R} \rightarrow S$, $f(x) = \{x\}$.

If you show $\#S \le c$ then you can conclude $\#S = c$ invoking Bernstein's theorem.

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Let's prove the following: let $X = \{X_k,\ k \in \mathbb{N}\}$ be a family of subsets such that $\#X_i \le c$, $\ \forall i \in \mathbb{N}$, and let $V = \bigcup_{k\in \mathbb{N}}X_k$. Then $\#V \le c$.

Because $\#X_i \le c$, we have for each $i$ a surjective $f_i: \mathbb{R} \rightarrow X_i$. We define $g: \mathbb{N} \times \mathbb{R} \rightarrow V$, $g(n,x) = f_n(x)$.

$g$ is an surjective function: let $x \in V$, then $x \in X_i$ for some $i$, then we have $a \in \mathbb{R}$ such that $f_i(a) = x$ because $f_i$ is surjective. Then $g(i,a) = f_i(a) = x$, with $(i,a) \in \mathbb{N}\times\mathbb{R}$.

We conclude that $g$ is surjective.

Thus $\#V \le \#(\mathbb{N}\times\mathbb{R})$, this is, $\#V \le c$.

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Your set $S$ satisfies the hypotheses, so we have $\#S \le c$, and thus $\#S = c$.