Let $X$ be a compact metric space. Let $X^{deg} = \{x \in X: \{x\}$ is a connected component$\}$. A space $X$ is said to be almost totally disconnected if the set $X^{deg}$ is dense in $X$. A point $x \in X$ is called isolated if and there exists a neighborhood of $x$ which does not contain any other points of $X$.
Give an example of an almost totally disconnected compact metric space that has no isolated points and contains only one connected component that is not a singleton - a unit interval.
My idea: Consider a unit square. Take unit interval over the number $0$ on $x$-axis. Take the classic $1/3$ Cantor set over the number $1$. Take Cantor sets over $1/2$ such that instead of middle third we remove middle ninth ($3^2$th), over $1/3$ such that we remove middle $3^3$th and so on.
I'm not sure if this works and I struggle to prove the desired properties.
Let $C(1)$ be the Cantor set. For $1<n\in \Bbb Z^+$ let $C(n)$ be a "fat Cantor set" with Lebesgue measure $1-1/n.$ Each $C(n)$ is an uncountable closed nowhere-dense subspace of $[0,1]$ with no isolated points.
Let $X=A\cup B$ where $A=[0,1]\times \{0\}$ and $B=\cup_{n\in\Bbb Z^+}(C(n)\times \{1/n\}).$