Rotten apple

Consider first a $2d$ version of the problem. Let $\Gamma$ be the circle delimiting the disc $\Delta$ on which the worm moves, $A∈\Gamma$ and $B∈\Gamma$ be respectively the starting and the end point of the trajectory of the worm and $[CD]$ be a diameter of $\Gamma$ . Suppose that the trajectory $\tilde{AB}$ intersects $[CD]$ in two points $E$ and $F$ such that $O$ the center of $\Gamma$ lies in $[EF]$ , then the length of $\tilde{AB}$ is at least equal to the diameter of $\Gamma$ . To prove this, assume WLG that $E$ and $F$ lie respectively in $[OC]$ and $[OD]$ , and denote by $\bar{XY}$ the length of the portion $\tilde{XY}$ of $\tilde{AB}$ , we use the fact that for a point $M$ on $\Delta$ , the nearest point on $\Gamma$ to $M$ is the intersection of $\Gamma$ and $[OM)$ , and the fact that the shortest path between two points is the straight line connecting the two points, to derive the inequalities $\bar{AE}≥CE$ , $\bar{EF}≥EF$ and $\bar{FB}≥FD$ , thus $\bar{AB}=\bar{AE}+\bar{EF}+\bar{FB}≥CE+EF+FD=CD$ . It follows that either $[OC]$ or $[OD]$ does not intersect the trajectory. Assume WLG that $[OC]$ does not intersect the trajectory. Let $V$ be a variable point on $\Gamma$ whose starting position is $C$ . Imagine that we move $V$ (for ex. clockwise), let $P$ be the position of $V$ when $[OV]$ touches $\tilde{AB}$ for the first time. Obviously $\tilde{AB}$ is completely outside the region delimited by $[OC]$ , $[OP]$ and the arc $\tilde{CP}$ (that corresponds to the clockwise sense starting from $C$ ). Let $Q$ be the 2nd intersection between $[PO)$ and $\Gamma$ , according to the first result the trajectory does not intersect $[OQ]$ , thus $[PQ]$ is a cut that satisfies the conditions of the $2d$ version. For the $3d$ version, we keep the same notations as before. Let $\Delta$ be the intersection of the ball and a plane containing $A$ , $B$ and $O$ the center of the ball. Consider the orthogonal projection of $\tilde{AB}$ on $\Delta$ , and let $[PQ]$ be a cut on $\Delta$ that satisfies the conditions of the $2d$ version. Consider the plane containing $[PQ]$ and perpendicular to $\Delta$ , this plane defines a cut that satisfies the condition of the $3d$ version.