To be a bit more formal: If I am choosing to disprove proposition $P$ is false, by exhibiting a counterexample, I would be proving.
"The proposition $P$ is false."
And the proof would start from
"If $P$, then for all integer (or whatever) $n$, $<$ some statement about $n$ that is implied by $P$ $>$". Therefore, in particular, for $n=$ my counterexample number, . This is a contradiction, so $P$ cannot be true.
In your simple example:
The proposition "The sum of any two integers is odd" is false.
If the sum of any two integers is odd, then (w) for all integers $a,b$, $a+b$ is odd. Consider in particular $a=1, b=1$. In that case $a+b=2$ which is not odd. This contradicts (w), thus our assumption that the sum of two integers is odd must be false.
PS - I have heard of a strong mathematician who excludes proof by contradiction from his axioms of what constitutes a valid proof. But I suspect that disproof by presenting a counterexample is still valid to him. In my formulation that would not work so there must be a better one.