Proving that if $a,b,c\in \mathbb N$ and $a^2+b^2=c^2$ then $abc$ is even. 
Let $a,b,c\in \mathbb N$ and $a^2+b^2=c^2$ then $abc$ is even.

My attempt:
If one or two numbers of $a,b,c$ are even then we're done, so we'll have to show that at least one of them is even. 
Suppose $a,b$ are odd, then $a^2,b^2$ are odd, so $a^2+b^2$ must be even. So $c^2$ is even then $c$ is even. So $abc$ is even.
Is this enough to prove it?
 A: Whether this is considered "enough to prove it" depends on your context. 
Your chain of arguments is perfectly valid. 
It is however possible you need to provide additional justifcation for the intermediate steps.
To get the idea across more clearly, suppose somebody sets the probem: "Show that  $a^2$ is odd if $a$ is odd." Then you could not just reply "It is." which is what you do effectively in you proof.
To be clear, it is fine, and even inevitable, to use results that are already known; all I mean to say is that you should check if you are allowed to use the results you use in your context. (This is something only you can do.) If this is the case, the proof is complete. If not, then not. 
A: Your proof is fine.
Here is a variation that perhaps is simpler.
We want to prove that at least one of $a,b,c$ is even, or equivalently, that they cannot all be odd.
Now consider these:


*

*$x$ odd implies $x^2$

*$x,y$ odd implies $x+y$ even
These imply that $a,b,c$ cannot all be odd: If they were, we'd get odd $+$ odd  $=$ odd.
A: Notice that a square of an integer is either $0$ or $1$ modulo $4$ therefore. If $$c^2\equiv0\mod{4}\Rightarrow c\equiv0\mod{2}\Rightarrow abc\equiv0\mod{2}$$
If $$c^2\equiv1\mod{4}\Rightarrow a^2\equiv0, b^2\equiv1\mod{4}\Rightarrow a\equiv0\mod{2}\Rightarrow abc\equiv0\mod{2}$$
or
$$a^2\equiv1, b^2\equiv0\mod{4}\Rightarrow b\equiv0\mod{2}\Rightarrow abc\equiv0\mod{2}$$
