Proving $2^{\mathbb N}$ is uncountable I know that there is a way to prove this and I am trying to figure this out using Cantor's diagonal method. 
For the most part, I understand how it works somewhat. You basically try to prove that there is a number that you can't get with the table or something related. You basically try to prove that infinite numbers are not one to one otherwise, you could simply count it. 
How can you all explain it better to me and use this to prove that $2^{\mathbb N}$ is uncountable?
 A: $2^{\mathbf{N}}$ means the set of all subsets of $\mathbf{N}$. Assume that the set is countable. Then there is an enumeration $A_0, A_1, A_2, \dots$, of all subsets of $\mathbf{N}$. 
You will obtain a contradiction if you can construct a subset $A$ of $\mathbf{N}$ that can't possibly be in the list $A_0, A_1, A_2, \dots$. Here is how you can construct such a set. Define $A$ so that, first, it disagrees with $A_0$ as to whether $0$ should belong to it or not. That is, if $0$ belongs to $A_0$, then define $A$ so that $0 \not\in A$; if $0$ doesn't belong to $A_0$, then define $A$ so that $0 \in A$. Next, decide whether $1$ will belong to $A$ in such a way that $A$ disagrees with $A_1$. And continue this way for $2$, $3$, etc. 
In general then, you have defined $A$ so that it is the set of all natural numbers $n$ for which $n \not\in A_n$. Since $A$ disagrees with each of the sets $A_0, A_1, A_2, \dots$, it can't be any one of them. Thus you have constructed a set which cannot be anywhere in the list, which is a contradiction.
