Let $A$ & $B$ be subsets of the real plane. Show that if $A$ is convex and is contained in $B$, which is a bounded set, then the length of the border of $A$ is $\leq$ than that of $B$.
The proofs of 2 and 3 are easier to carry out yourself than to watch/read someone else do it.
First, show that for a general half-space $H$, $B\cap H$ has smaller boundary than $B$.
If $A$ is sufficiently smooth, you can write $A$ as the intersection of $B$ with a countable family of half-spaces, say
From this (and lower semicontinuity of the perimeter functional, see any text of sets of finite perimeter), it follows that $A$ has smaller boundary than $B$.