Find the cardinal of $X=\{A\subseteq \Bbb N : A \text{ is finite } \}$ Find the cardinal of $X=\{A\subseteq \Bbb N : A \text{ is finite } \}$.
Attempt
$$ X=\bigcup_{i\in\Bbb N}X_i$$
Where $X_i$ denotes the set of subsets of $\Bbb N$ with cardinal $i$.
Let $A\in X_i, A=\{a_1,a_2,...,a_i\}$. I define the functions $f_i:X_i\to\Bbb N^i,f_i(A)=(a_0,a_1,...,a_i)$ (assume the elements are ordered when putting $A$ as an argument of $f_i$, to secure the welldefinedness).
These functions are obviously injective, thus $|X_i|\leq \aleph_0^i=\aleph_0$. 
We also have the (trivial injection) $g:\Bbb N\to X_1:g(n)=\{n\}$.
Thus we get $|X_i|=\aleph_0$, and as $X$ is a countable union of countable sets we can conclude that $|X|=\aleph_0$.
 A: Your proof looks fine, but you can come up with an explicit injection (indeed, it's probably a bijection, but I can live without one of those) $h:X \to \mathbb{N}$ as follows: for each $A$, take the sequence $(a_i)_i$ which is $1$ when $i \in A$ and $0$ otherwise. This sequence has a finite number of $1$s. Hence,
$$ h(A) = \sum_i a_i 2^i $$
converges, since the sum is finite. It is an injection as you can read off the set essentially from the (unique) binary expansion of the integer $h(A)$: $2^m$ is in this if and only if $m \in A$.
Edit: Okay, let's have an example. Starting with $h(A)$, we shall recover $A$. $h(A)=10011110$ is the unique binary expansion of some integer (158, as it happens, but this is irrelevant). This is shorthand for
$$ 0 \cdot 2^0 + 1 \cdot 2^1 + 1 \cdot 2^2 + 1 \cdot 2^3 + 1 \cdot 2^4 + 0 \cdot 2^5 + 0 \cdot 2^6 + 0 \cdot 2^7 + 1 \cdot 2^8 + 0 \cdot 2^9 + \dotsb, $$
where the highest power of $2$ which is multiplied by a $1$ is $2^8$. We find the set $A$ by writing down all the exponents of the powers of two which are multiplied by $1$ in this expansion, so in this case, the set is
$$ A=\{ 1,2,3,4,8 \}. $$
One can show that $h$ is injective directly: suppose that $A,B$, $A \neq B$ are two sets in $X$. WLOG there is a natural number $a \in A$, $a \notin B$ (if not, just swap $A$ and $B$: if there is no such number, $A=B$). Therefore under the map $h$, $A$ maps to some binary integer $\alpha$, in which the $a$th digit from the right (counting the initial one as $0$th) is $1$, whereas $B$ maps to a binary integer with a zero as the $a$th digit. Therefore $A \neq B \implies h(A) \neq h(B)$, and we have an injection. (It is easy to see that the inverse is an injection in exactly the same way.)
(Oh, and by the way, the terminology is usually that we have the cardinals as a class of objects, and sets in bijection with the cardinal $\alpha$ are said to have cardinality $\alpha$.)
