Consider the set $X=\left\{(x_n)_{n\in\mathbb{N}}\mid x_n\in\{0,1\},\forall n\in\mathbb{N}\wedge x_n=1\text{ for at most finitely many $n$}\right\}$ Then what can we say about the cardinality of $X$ (countable or uncountable)?
My try: I've taken an arbitrary $A=\{s_1,s_2,\ldots,s_n,\ldots\}\subseteq X$ which is countable.
Arrange the sets \begin{align}Y_0&=\{(s_k)_{k\in\mathbb{N}}\mid s_k\text{ contains no }1's\}\\ Y_1&=\{(s_k)_{k\in\mathbb{N}}\mid s_k\text{ contains exactly one }1's\}\\\vdots \\ Y_m&=\{(s_k)_{k\in\mathbb{N}}\mid s_k\text{ contains exactly $m$ 1's}\}\\\end{align}
This process must stop for some $m\in\mathbb N$ because each sequence of $X$ contains at most finitely many $1.$
Clearly $A=\bigcup_{i=0}^mY_i$
So if we consider the set $S=\{(x_n)_{n\in\mathbb{N}}\mid x_n\in\{0,1\}\;\forall n\in\mathbb N\text{ and $x_n=1$ for $(m+1)$ values of $n$} \}$
Then $S\not\in A.$ $A$ was an arbitrary countable subset of $X$ and we've shown that $A$ is a proper subset of $X$. Thus any countable subset of $X$ is a proper subset of $X$.
If $X$ is countable then according to the proof, $X$ being a countable subset of $X$, is a proper subset of $X$, a contradiction. Hence $X$ must be uncountable.
$\mathcal{P}_{fin}(\mathbb{N})$ (i.e. the collection of the finite subsets of $\mathbb{N}$) is countable. Clearly $\mathbb{N},\mathbb{N}^{(2)},\mathbb{N}^{(3)},\ldots$ are countable, so each element of $\mathcal{P}_{fin}(\mathbb{N})$ can be identified with a couple of natural numbers: the first natural number is the number $k$ of elements of such subset and the second number is the index of such subset among the elements of $\mathbb{N}^{(k)}$. It follows that $$ \left|\mathcal{P}_{fin}(\mathbb{N})\right| = \left|\mathbb{N}\times\mathbb{N}\right|=|\mathbb{N}|=\aleph_0.$$