Let $C$ be a closed polygon in $\mathbb{R}^2$. I would like to find a maximal convex figure contained in $C$. "Maximal" means that it is not contained in any other convex figure contained in $C$ (it does not need to have maximum area).
I know how to find a maximal axes-parallel rectangle contained in $C$: just start from an arbitrary point inside $C$, and stretch it to the right and left and top and bottom, until it bumps into the boundary of $C$. This process is well-defined since $C$ is closed (contains its boundary).
But, finding a maximal convex figure seems more complicated since there are infinitely many directions to stretch...
Let $P$ be an solid polygon (polygon $\cup$ inside): Pick $x \in P$ and Define $$ \Gamma=\{ A \subseteq P ~| ~x\in A , ~ A \text{ is nonempty convex} \} $$ Then the set $\Gamma \neq \emptyset$ since $\{x\} \in \Gamma$ and it is partially ordered under inclusion $(\subseteq).$ Now picking any chain in $\Gamma$, say $\{A_{\alpha}\}_{\alpha \in I},$ one can show that $\bigcup_{\alpha \in I} A_{\alpha} \in \Gamma $. So $\{A_{\alpha}\}_{\alpha \in I},$ has an upper bound in $\Gamma,$ and therefore Zorn lemma guarantees $\Gamma$ has a maximal element, say $M$, it is actually a maximal convex subset of $P$ containing $x.$ Thus you pick $x \in P$ and start convex expansion to get one maximal convex subset. (It is always possible based above argument!)
Note that Since $P$ is bounded this expansion is bounded, you finally stop somewhere in convex expansion process. And since Polygon has a finitely many extreme points, the number of such maximal convex components is finite .