Suppose such that . Then there exists a neighborhood of such that for all . (Prove this using the continuity of at .) Then which is false.
I don't see how anything that you mention really has much to do with the problem. The solution I gave is quite standard : suppose the function is not identically zero, then it's bounded away from zero on a small subinterval, and hence its integral on this small subinterval is positive, and the integral on the whole interval is at least the integral on this small subinterval.