How do I write this proof down? The exercise:
$A = \{ n \in \mathbb{N} : n = 4 m ^{2} \text{ for certain } m \in \mathbb{N}\},$
$B = \{ n \in \mathbb{N} : n  \text{ is even} \},$
$C = \{ n \in \mathbb{Z} : n = m ^{2} \text{ for certain } m \in \mathbb{Z}\}.$
now prove that
$A \subseteq B \cap C$
The proof:
If we have $n$ that is in $A$, it has divisor $2$ and must be in $B$ as well.
If we state that for each $m$ $4m^{2}=n^{2}$ for $n=2m$ then $A$ must be in $C$ as well. If $n$ that is in $A$ is in both $B$ and $C$, than it is proven. How do I correctly write this down?
 A: You've pretty much got it. It suffices to show that, for any arbitrary $n\in A$, we have both $n\in B$ and $n\in C$. If you want to prove equality, you need to show that if $n\in B$ and $n\in C$, then we also have $n\in A$.
A: Take $n\in A$. Then by definition of $A$, there exists $m\in\mathbb N$ such that $n=4m^2=2*2m^2=(2m)^2$. Setting $h=2m$, one has $h\in\mathbb Z$ and $n=h^2$, therefore $n\in C$; moreover $n=2*(2m^2)$ is even, hence $n\in B$. By arbitrariness of $n\in A$, we conclude $A\subseteq B\cap C$.
A: Your proof is (pretty much) correct. I would say that your proof is written dowb already. But since you ask about how to write it down, I will give you my thoughts. 
When I have problems with writing a proof down, I "try" to go through everything in detail. I try to write down as much that I can and I try to make sure that everything that I do write down is absolutely correct and that I understand ever part of the proof. If something is unclear, I rewrite it. I try to use a lot of text to explain my thoughts. Often, I think, people have the right thoughts in the head, but they "forget" to write them down. They just assume that the reader is making the same connection that you are. Anyway, my approach usually means that the proofs are longer than necessary, but it also means that I have a better understanding of the proof. And that understand is what allows me to shorten the proof. So you might think of the longer version as a type of draft.
You could write the proof like this:
Proof: I want to show that $A \subseteq B\cap B$. That means that I need to show that every element in $A$ is also an element in $B$ and $C$. 
So let now $n\in A$. That $n \in A$ means that $n = 4m^2$ for some $m\in \mathbb{Z}$. Since the product of an even number with any other number is always even, $n$ must be even. But that implies that $n \in B$. 
Let us then show that $n\in C$. Now again, as above, we have $n = 4m^2$ for some $m\in \mathbb{Z}$. So we have $n = 4m^2 = (2m)^2$. Then indeed we have $n = k^2$ ($k = 2m$) and this implies that $n\in C$. $\square$
Again, this is way too much detail, but maybe you can shorten the proof.
