Can there be maximal proper subcontinua? Let X be a compact connected Hausdorff space (a continuum).
Say $C$ is a proper subcontinuum: a proper compact connected subspace of $X$.
Can $C$ be maximal? That is: is it possible that for all proper subcontinua $C \subset C' \subset X$, we have that $C = C'$?
What about under the stronger condition that $C \subset \text{int}(C')$?
I can prove that if $C$ is maximal then every point outside of $C$ is a non-cut-point. And in every example I can think of there cannot be maximal proper subcontinua, so I think they don't exist.
If there isn't an example and I should be able to prove this a hint will suffice. If there is an example I would like to see it though. I am not so good at coming up with examples yet.
 A: No, this is impossibile.
Here is a simple way of proving this through the boundary bumping theorem (there is also an argument looking at order arcs in the hyperspace $C(X)$ of subcontinua with the Vietoris topology). First recall the boundary bumping theorem:
Theorem (boundary bumping): Let $K$ be an Hausdorff continuum and let $O\subset K$ be open. If $D$ is a connected component of $\overline{O}$, then $D\cap\partial O\neq\varnothing$.
Now let's prove the result you need. Let $C\subseteq X$ be a proper subcontinuum and find an open $U$ such that $C\subseteq U\subseteq X$ and $U\neq X$ (this can be done by normality, separating $C$ and a point in $X\setminus C$ for example). Now by normality again we can find a second open set $V$ such that $C\subseteq V\subseteq\overline{V}\subseteq U$. Consider as $C'$ the connected component of $\overline{V}$ containing $C$ (this exists since $C$ is connected). By the boundary bumping theorem $C'\cap\partial V\neq \varnothing$, in other words $C'\cap(X\setminus V)\neq\varnothing$, which implies that $C'$ while still being a proper subcontinuum is strictly bigger than $C$, so $C$ is not maximal.
