I think it is something with the derivative... f'(x)=|f(x)-f(y)|/|x-y|=0 when epsilon is very small.
f'(x)=0 for all x ==> f(x) constant for all x.
Hmmm...
You could state this a little bit more rigorously.
1) Let then the above says that . Thus, since we have that (by continuity of ) and thus by the squeeze theorem is differentiable at and . Since was arbitrary it follows that for all
2) Since is an interval we may conclude that .
Just a comment though.