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.