Claim: if $a, b, c \in \mathbb{N}$, then at least one of $a-b$, $a+c$, $b-c$ is even.
Verdict: this is a true statement.
Proof: (Contradiction) Suppose that there isn't one of $a-b$, $a+c$, $b-c$ that is even. Thus, assume that $a-b$, $a+c$, $b-c$ are all odd integers. We write $$S_1: a-b=2x+1, x \in \mathbb{Z}$$ $$S_2: a+c=2y+1, y \in \mathbb{Z}$$ $$S_3: b-c=2z+1, z \in \mathbb{Z}$$ by the definition of an odd integer (where $S_n$ represents the $n$th statement).
- Supplementary proposition: if $j,k \in Z$ have different parity (that is, either $j$ must be odd and $k$ even or $j$ must be even and $k$ odd), then $j+k$ and $j-k$ is odd.
Thus $a$ and $b$ in $S_1$, $a$ and $c$ in $S_2$ and $b$ and $c$ in $S_3$ must have different parity. However, if we take $a$ in $S_1$ to be an even integer, $b$ must be an odd integer by the supplementary proposition and our initial assumption that $a-b$, $a+c$ and $b-c$ are all odd. But since $a$ is an even integer, $c$ must be an odd integer in $S_2$ by the same reasoning. We have a contradiction in $S_3$; that is, $b$ is odd and $c$ is odd, which violates violates our proposition and, in effect, contradicts our assumption that $a-b$, $a+c$ and $b-c$ are all odd integers. Thus, at least one of $a-b$, $a+c$, $b-c$ must be even. $ \square$
What you have is fine as far as it goes, but to complete the argument you need to consider also the case of odd $a$.
A simpler approach is to notice that
$$(a-b)+(a+c)+(b-c)=2a\;,$$
which is even. The sum of three odd numbers, however, is odd, so at least one of the three must be even. (In fact either all three are even, or exactly one is even; you should try to prove this.)