 Solution

Suppose, for a contradiction, that $f$ has no zero in the open disk. The strict boundary inequality implies no boundary zero either. Thus $1/f$ is holomorphic on a neighborhood of the closed disk: compactness and continuity keep $f$ nonzero in a sufficiently small neighborhood. Apply the $k=0$ Cauchy estimate to $1/f$:This contradicts the assumed strict inequality. If $f(z_0)=0$, the conclusion is immediate without taking a reciprocal. Therefore $f$ has a zero in the open disk. This is a local form of the minimum modulus principle.