I am wondering if given a separable metric space $X$, it is possible to totally disconnect $X$ by repeatedly removing countable dense subsets.
For example, let $I_1$ be a countable dense subset of $X$, and let $Y_1=X\setminus I_1$. Inductively, define $I_n$ as a countable dense subset of $Y_{n-1}$ and $Y_n=Y_{n-1}\setminus I_n$.
My question is, what happens eventually? Is there some $n\in\mathbb{N}$ for which $Y_n$ is totally disconnected, and if so how would one go about proving it? If not, are there any other relevant results associated with this? What about $Y_\infty:=\bigcap_{i=1}^\infty Y_i=X\setminus (\bigcup_{i=1}^\infty I_i)$ ?
$(X\setminus I_1)\setminus I_2=X\setminus(I_1\cup I_2)$. The countable union of countable dense sets is still countable dense. So all the sets $I_n$ can be joined into one countable dense set. So the question asks whether a separable metric space is the union of a totally disconnected space and a countable dense set.
For $\mathbb{R}$, this is true: $\mathbb{Q}$ is the countable dense set, $\mathbb{Q}^c$ is totally disconnected.
For $\mathbb{R}^2$, it is false. $\mathbb{R}^2\setminus C$ for $C$ countable is still connected, in fact path-connected. At any of its points $p$, the set of lines through $p$ is uncountable, whereas the lines joining $p$ to points in $C$ is countable; so $\mathbb{R}^2\setminus C$ is full of intersecting lines.