That's a standard inequality, though perhaps not well enough known for the usual "take it to stackexchange" comment.  Denote by $\partial \Sigma$ the boundary of any set $\Sigma$ in the plane.  In the case that $s$ is a polygon we can use induction on the number $k$ of edges of $\partial s$ not contained in the $\partial S$.  If $k=0$ we're done.  If $k>0$, choose an edge $e$ of $s$ not in $\partial S$, and let $H$ be the closed half-plane such that $H \supset s$ and $\partial H \supset e$.  Then $S' := S \cap H$ is a convex set containing $s$ whose boundary is shorter than $\partial S$ because we've replaced part of $\partial S$ with the line segment joining the same two points.  Moreover $\partial s$ has $k-1$ edges not contained in $\partial S'$.  This completes the induction step and the proof.

That argument applies more generally when $\partial s \setminus \partial S$ is polygonal.  If it's not, we can reduce to that case via a limiting argument, replacing each component of $\partial s \setminus \partial S$ by an arbitrarily close polygonal path.