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2 Changed $\delta$ to $\partial$ for consistency; corrected transposition typo in "arbitrarily"

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 $\delta \partial s$ has $k-1$ edges not contained in $\delta \partial S'$. This completes the induction step and the proof.

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

1

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 $\delta s$ has $k-1$ edges not contained in $\delta S'$. This completes the induction step and the proof.

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