If $a,b,c\in\Bbb N$ then at least one of $a-b$, $a+c$, and $b-c$ is even.
For the sake of contradiction, assume $a-b$, $a+c$, and $b-c$ are all odd. This gives us $a-b=2l+1$, $a+c=2m+1$, and $b-c=2n+1$ for $l,m,n\in\Bbb N$.
Now we have:
$\begin{align*}a-b&=2l+1\\a&=2l+b+1\end{align*}$
Substituting:
$\begin{align*}(2l+b+1)+c&=2m+1\\2l+b+c&=2m\\b&=2m-2l-c\end{align*}$
Substituting again:
$\begin{align*}(2m-2l-c)-c&=2n+1\\2m-2l-2c&=2n+1\\2c&=2m-2l-2n-1\\c&=m-l-n-\frac12\end{align*}$
But this is impossible for $c\in\Bbb N$, thus a contradiction and we are finished.
$\blacksquare$
I'm wondering if my reasoning is sound for this proof.. is there a simpler way I could have done it? Any advice welcome
Your proof is fine.
Here's any easier one: Suppose $a+c$, $a-b$, $b-c$ are odd. Then their sum is odd. This is a contradiction since there sum is $2a$, which is even.