Let $A$ be the set of all finite subsets of $\mathbb{Z}_+$, then is $A$ countable?
Defiene $I_j$ as the set consisting of all subsets of $\mathbb{Z}_+$ having $j$ elements, then i think $A=\bigcup I_j$ and since countable union of countable sets are countable thus $A$ is countable!
Let $p_1,p_2,p_3,\dots$ be the primes listed in increasing order.
Let $A$ be a finite set of positive integers. For any positive integer $k$, let $A(k)=1$ if $k\in A$, and let $A(k)=0$ otherwise. Let $$\varphi(A)=\prod_{k=0}^\infty p_k^{A(k)}.$$ The product above is a positive integer, since $A$ is finite. By the Unique Factorization Theorem, the function $\varphi$ is one to one.
The function $\varphi$ is a one to one function from the set of finite subsets of the positive integers to the positive integers. It follows that the set of finite subsets of the positive integers is countably infinite.
Here's another way to do it: for any finite $I \subset \Bbb Z_+$, list the elements in increasing order; thus we may write $I = \{i_1, i_2, i_3, . . ., i_k \}$, where $\#(I) = k$ (here I use the "$\#$" symbol to denote cardinality). Let $p_j$ be the $j$-th prime, so that $p_1 = 2$, $p_2 = 3$, $p_3 = 5$, and so forth. Assign the natural number $\prod_1^k p_j^{i_j}$ to I; call the function so defined $\phi$. Thus $\phi:A \to \Bbb Z_+$. The key is to note that $\phi$ is injective. The injectivity of $\phi$ follows from unique factorization into primes. Thus we have a one-to-one function $\phi$ which assigns to each $I \in A$ a unique natural number, effectively enumerating $A$. QED.
And I thank you, as does Dr. Godel! ;)
NOTA BENE: Answer edited to remove false assertion that $\phi$ surjective. $\phi$ cannot in fact be surjective since in fact the order of the $i_k \in %$ corresponds to the order of the primes; thus, for example, $2^5 3^7$ represents the set $\{5, 7\}$, but $2^7 3^5$ is not the image of any $I \in A$ under $\phi$. "Haste Makes Waste," is my new motto!
