The cardinality is at least $|X|$, since $S$ contains all singletons.
Let $S_n$ be the subset of $S$ consisting of all subsets of cardinality exactly $n$. Then $S$ is the disjoint union of the $S_n$.
Now, for any positive integer $n$, the number of subsets of $X$ of cardinality $n$ is at most $|X|^n = |X|$ (equality since $|X|$ is infinite); because an $n$-tuple of elements of $X$ determines a subset of $X$ of cardinality at most $n$; and every subset with $n$ elements determines only finitely many distinct $n$-tuples of elements of $X$ (namely, $n!$). So $|S_n|\leq n!|X|^n = |X|$ for all $n$.
Therefore:
\begin{align*}
|X| &\leq |S| = \left|\bigcup_{n=0}^{\infty} S_n\right| = \sum_{n=0}^{\infty}|S_n|\\
&= \sum_{n=1}^{\infty}|S_n| \leq \sum_{n=1}^{\infty}|X|\\
&= \aleph_0|X| = |X|,
\end{align*}
with the last equality since $|X|\geq\aleph_0$.
Thus, $|X|\leq |S|\leq |X|$, so $|S|=|X|$.
It is indeed $2^{\aleph_0}$. $\Bbb N$ has $2^{\aleph_0}$ subsets altogether, so certainly it has no more than $2^{\aleph_0}$ infinite subsets. On the other hand, there is an injection $\varphi$ from $\wp(\Bbb N)$ into the set of infinite subsets of $\Bbb N$.
Let $O=\{2n+1:n\in\Bbb N\}$, the set of odd positive integers. Then for each set $S\subseteq\Bbb N$ let $$\varphi(S)=O\cup\{2n:n\in S\}\;.$$
Since $\varphi(S)\supseteq O$ for every $S\subseteq\Bbb N$, every $\varphi(S)$ is infinite. On the other hand, it’s clear that $\varphi$ is injective (one-to-one), since whenever $S_0,S_1\subseteq\Bbb N$ with $S_0\ne S_1$, $\varphi(S_0)\setminus O\ne\varphi(S_1)\setminus O$, and therefore $\varphi(S_0)\ne\varphi(S_1)$.
Intuitively, $\varphi(S)\setminus O$ looks just like $S$, except that it’s been blown up by a factor of $2$: it’s simply the double of $S$. Thus $S$ is completely recoverable from the even numbers in $\varphi(S)$: just divide each of them by $2$. This ensures that $\varphi$ is an injection. Throwing in $O$ just ensures that $\varphi(S)$ is always infinite.
Thus, if $\mathscr{A}$ is the set of infinite subsets of $\Bbb N$, $\varphi$ is an injection of $\wp(\Bbb N)$ into $\mathscr{A}$, while the identity map is an injection of $\mathscr{A}$ into $\wp(\Bbb N)$, and it follows from the Cantor-Schröder-Bernstein theorem that $|\mathscr{A}|=|\wp(\Bbb N)|=2^{\aleph_0}$.
Best Answer
Suppose that the set of all infinite subsets (and thus all are countable) of $\mathbb N$ was countable. Suppose thus that $\{A_n\}_{n\in \mathbb N}$ is a list of all the countable subsets of $\mathbb N$, and write $A_n=\{a_{n,1},\cdots, a_{n,k},\cdots \}$. Now diagonalize by defining $b_n$ to be some integer different than $\{b_j\}_{1\le j < n}$ and also $b_n\ne a_{n,n}$ and show that $\{b_n\}_{n\in \mathbb N}$ is not on your list, but it is an infinite subset of $\mathbb N$.