How many infinite subsets does $\mathbb N$ have? I attempted to write a derivation of the answer, but was told my mathematics was wrong; please correct me.           
The cardinality of $\mathbb N$ is $\aleph_0$.
From this set, we can generate another infinite subset by excluding $1$ element.
There are $\aleph_0$ such possible subsets that can be generated like this.
We can generate an infinite subset by excluding $2$ elements from $\mathbb N$.
There are $\aleph_0 \choose 2$ possible subsets that can be generated like this.   
In general, for any $i$ from $0$ to $\aleph_0$ we can generate $\aleph_0 \choose i$ such possible subsets by excluding $i$. To find the total number of possible subsets, we simply sum all the combinations.
$$\sum_{i = 0}^{n} {n \choose i} = 2^n$$    
Based on the above:
$$\sum_{i = 0}^{\aleph_0} {\aleph_0 \choose i} = 2^{\aleph_0}$$    
$2^{\aleph_0} = \aleph_1$
$\therefore$  the number of infinite subsets of $\mathbb N$ is $\aleph_1$.   
I realise that I excluded the number of infinite subsets who have infinite complements.     
To account for this, merely combine any $k$ $i$ used in the selection above, and exclude all multiples of the products of $i_1*i_2*i_3*...*i_k$.
We have $\aleph_0$ such sets of $i$ with numbers increasing from $0$ to $\aleph_o$.    
I didn't consider this when I first wrote it out, and only realised it after. I haven't yet updated my proof to include it. However, this wasn't the problem with my proof; I was told I did "bad mathematics".
 A: $\Bbb N$ has $2^{\aleph_0}$ subsets. The number of subsets of size $n$ for any $n$ finite is countable. The union of a countable number of countable sets is countable, so $\Bbb N$ has a countable number of finite sets...
A: Some of the math you did (plugging cardinalities into the combination function for instance) isn't valid. The function $(n,k)\mapsto {n\choose k}$ takes nonnegative integers as arguments; the expression ${\aleph_0 \choose n}$ is meaningless (if such a thing has been defined and is useful, it's probably the case that you haven't learned about it in your class). 
Let $X$ be the set of infinite subsets of $\mathbb{N}$. To prove the claim, note that $\vert \mathscr{P}(\mathbb N)\vert =\mathfrak{c}$. Further, there are $\aleph_0$ finite subsets of $\mathbb N$ (since the set of all finite subsets of $\mathbb N$ is a countable union of countable sets). Therefore,
$$\aleph_0+\vert X\vert = \mathfrak c$$
Since $\aleph_0+\alpha=\alpha$ for any infinite cardinal $\alpha$, we have $\vert X \vert = \mathfrak c$. 
