 Solution

Let $c$ be the center of the containing disk and take arclength around the closed curve. The periodic function $\Phi (s)=|\gamma (s)-c{|}^2/2$ has a maximum. At a maximum $s_0$,By the Cauchy-Schwarz inequality and the containing radius, $1\le |\gamma (s_0)-c||{\gamma }^{\prime \prime }(s_0)|\le r|k(s_0)|$. ThereforeClosedness provides an interior maximum on the periodic parameter domain. Without it, the straight segment $\gamma (s)=(s,0)$ for $-r<s<r$ lies in the disk but has signed curvature identically zero, giving the required counterexample.