Assume that . You know that f takes on a minimum on . Say it takes on its minimum at . Note that since .
, contradicting the fact that is the minimum of f on I.
Additionally, you could suppose that on the interval (if you can't do this then f is the zero function and there is nothing to prove).
Then choose a point , so you have a new point such that .
The by induction you have points which satisfy
Now some subsequence converges to a point . And since is continuous
.
Really, the wasn't important. You just needed a constant and the same proof will work.
f has a root in [a,b] if and only if |f| has a root in [a,b], right?
If |f| doesn't have a root in [a,b], then 0 < |f(x)| for all x in [a,b], right? From that assumption, I derived a contradiction. Which must mean that there indeed does exist a point x = c such that 0 = |f(c)| = f(c)