Example of an $\omega_1$ decreasing chain of dense semicontinua? In his well-known paper Bellamy constructs an indecomposable continua with exactly two composants. The setup is as follows:
We have an inverse-system $\{X(\alpha); f^\alpha_\beta: \beta,\alpha < \omega_1\}$ of metric indecomposable continua and retractions. For each $X(\alpha)$ a composant $C(\alpha) \subset X(\alpha)$ is specified and each $C(\alpha)$ maps into $C(\beta) $. 
The inverse limit $X$ has exactly two composants. The first is the union
$\bigcup\{X(\beta): \beta < \omega_1\}$ where we identify $X(\beta)$ with the set of sequences $(x_\alpha) \in X$ with $x_\beta = x_{\beta+1} = x_{\beta+2} = \ldots . $ The second composant is the inverse limit $\{C(\alpha); f^\alpha_\beta: \beta,\alpha < \omega_1\}$.
Observe there is no reason a priori for the second composant to be nonempty. However I do not believe an example is know.
My question is an easier one. Can you think of an example of a metric continuum $M$ and a $\omega_1$-indexed decreasing nest of dense semicontinua with empty intersection? We call the set $S \subset M$ a semicontinuum to mean for each $x,y \in S$ there exists a continuum $K$ with $\{x,y\} \subset K \subset S$.
If the second composant was empty the family $\{f^\alpha_0(C(\alpha)): \alpha < \omega_1\}$ would be such a nest for $M = X_0$.
If we index by $\omega$ instead an example is easy to come by. Let $M$ be the unit disc and $Q = \{q_1,q_2, \ldots\}$  and enumeration of the rational points on the boundary. Let $S(n)$ be formed by drawing the straight line segment from each element of $\{q_n,q_{n+1}, \ldots\}$ to each rational point of $(0,1/n) \times \{0\}$. Then add in $(0,1/n) \times \{0\}$ itself to make the space a semicontinuum.
Indexing by $\omega_1$ must somehow get around the fact that any $\omega_1$ decreasing nest of closed subsets of a metric continuum eventually stabilizes.
It feels like this would be easier if we assume the Continuum Hypothesis.
 A: Indeed, if CH holds then there is an enumeration $\{x_\alpha:\alpha<\omega_1\}$ of all the points in the square $M:=[0,1]^2$.  Then take $S(\alpha)=M\setminus \{x_\beta:\beta<\alpha\}$. 
Each $S(\alpha)$ has countable complement and is therefore path/arc-connected $-$ for every two points $x,y\in S(\alpha)$ there is actually an arc in $S(\alpha)$ consisting of two straight line segments which joins $x$ and $y$. So $S(\alpha)$ is a semicontinuum.
It is clear that $\overline{S(\alpha)}=M$, the $S(\alpha)$'s are decreasing, and $\bigcap\{S(\alpha):\alpha<\omega_1\}=\varnothing$. 
Side note: I thought this problem was interesting when I first studied Bellamy's example a couple years ago. I've also wanted to use his inverse limit of retractions method to find a metrizable decomposable continuum $X$ which has only finitely many (or countably many) equivalence classes $x/\sim$, where $x\sim y$ if and only if there is a nowhere dense subcontinuum of $X$ containing $x$ and $y$.  I believe this is also an open problem.
A: Let $\ \Omega\ := \{\alpha: \alpha<\omega_1\},\ $ and $\ i:\Omega\rightarrow [0;1]\ $ be injective. Let $\ M:=[0;1]^2.\ $ Then
$$ S_\alpha\ :=\ M\ \setminus \ i([0;\alpha))\times(0;1] $$
for $\ \alpha<\omega_1,\ $ is the required decreasing $\omega_1$-sequence

A VARIANT A similar example admits
$$ M:=[0;1]\times\{0\}\cup C\times[0;1]$$
This time continuum $\ M\ $ is $1$-dimensional (which feels thematic).
A: 
I am not assuming any CH.

Let $\ \Omega\ := \{\alpha: \alpha<\omega_1\},\ $ and $\ i:\Omega\rightarrow [0;1]\ $ be injective and such that
$\ \Gamma\ :=\ i(\Omega)\ $ is condensed in $\ [0;1].\ $ More generally,
let $\ \Gamma_\alpha\ :=\ i([0;\alpha)).\ $ Also let $\ M:=[0;1]^2.\ $
Then
$$ S_\alpha\,\ :=\,\ (\Gamma\setminus\Gamma_\alpha)\times[0;1]
\ \cup\ [0;1]\times(\Gamma\setminus\Gamma_\alpha) $$
for $\ \alpha<\omega_1,\ $ is the required decreasing $\omega_1$-sequence which has empty intersection.


Condensed means that every non-empty open subset of $\ [0;1]\ $ has an uncountable intersection with $\ \Gamma$.

