I will assume the Axiom of Choice throughout.
Let $X$ be any infinite set, with $|X|=\kappa$, and for any $n\in\mathbb{N}$, let $X^n$ be the set of ordered $n$-tuples.
There is a surjection from $X^n$ to the collection of all subsets of $X$ with at most $n$ elements, namely the map that takes the $n$-tuple $(x_1,\ldots,x_n)$ to the subset $\{x_1,\ldots,x_n\}$. That means that
$$\mathcal{P}_n(X) = \{A\subseteq X\mid |A|=n\}$$
has cardinality at most that of $X^n$.
But for infinite cardinals $\kappa$, $\kappa\times\kappa = \kappa$ if we assume the Axiom of Choice (in fact, this is equivalent to AC), from which it follows by induction that $\kappa^n = \kappa$ for all $n\in\mathbb{N}$. So $|\mathcal{P}_n(X)| \leq |X^n| = \kappa$. Moreover, $\kappa\leq |\mathcal{P}_n(X)|$ (biject $X$ with $X-\{x_1,\ldots,x_{n-1}\}$ for some subset of exactly $n-1$ elements to get that there are at least $|X|$ subsets with $n$ elements). So $|\mathcal{P}_n(X)| = |X|=\kappa$.
Thus, if we let
$$\mathcal{P}_{\lt\infty}(X) = \{A\subseteq X \mid |A|\lt\aleph_0\},$$
then
$$\begin{align*}
\left|\mathcal{P}_{\lt\infty}(X)\right| &= \left|\bigcup_{n\in\mathbb{N}}\mathcal{P}_n(X)\right|\\
&= \sum_{n\in\mathbb{N}}\left|\mathcal{P}_n(X)\right|\\
&\leq \sum_{n\in \mathbb{N}}\kappa\\
&= |\mathbb{N}|\kappa = \aleph_0\kappa = \kappa,
\end{align*}$$
with the last equality because if we assume the Axiom of Choice, and $\kappa$ and $\lambda$ are any two cardinals, at least one infinite and neither one equal to $0$, then $\kappa+\lambda = \kappa\lambda = \max\{\kappa,\lambda\}$.
Now, for each $A\in \mathcal{P}_{\lt\infty}(X)$, the subgroup of $S_X$ with support contained in $A$ has $n!$ elements, where $|A|=n$. Therefore,
$$\begin{align*}
|S_X| &\leq \sum_{A\in\mathcal{P}_{\lt\infty}(X)} |S_A|\\
&\leq \sum_{A\in\mathcal{P}_{\lt\infty}(X)} \aleph_0\\
&= |P_{\lt\infty}(X)|\aleph_0 = \kappa\aleph_0 = \kappa.
\end{align*}$$
On the other hand, if we fix $x\in X$ and consider all transpositions of the form $(x,y)$ with $y\in X-\{x\}$, then we have that $|S_X$ has at least $|X|=\kappa$ elements. Therefore, $|S_X| = \kappa$.
That is: if $X$ is infinite, then the group of permutations of $X$ with finite support has cardinality $|X|$.