When is the complement of a disjoint countable union of compact sets connected? Given that $X$ is a metric space, with disjoint compact subsets $C_1, C_2, ... \subseteq X$. Define $C := \bigcup_{k=1}^\infty C_k$.
I want to make the following claim:

If $C$ is bounded, then:
$X\setminus C$ is disconnected $\iff$ There exists $k \in \mathbb{N}^{\geq 1}$ such that $X \setminus C_k$ is disconnected

Now obviously this claim is not true in general (just take any countable set to be $X$ and use the discrete metric). However what happens in the case where $X = \mathbb{R}^n$ (with the standard metric)? Does anyone know how to attack this problem?
I feel like this result should be true in the above case, but if it's not then does it at least become true if we restrict the $C_k$ subsets to having to be connected topological manifolds?
 A: This is only a partial answer.
Note that $n = 1$ is a trivial case because $\mathbb R \setminus A$ is always disconnected if $A$ is a non-empty bounded set.
Let us prove that if $\mathbb R^n \setminus C_k$ is disconnected for some $k$, then also $\mathbb R^n \setminus C$ is disconnected.

*

*The countable union $D = \bigcup_i D_i$ of disjoint non-empty compact subsets $D_i \subset \mathbb R^n$ cannot be a connected open subset of $\mathbb R^n$.
Assume that $D$ is connected and open. The boundary $\operatorname{bd} D_1$ of $D_1$ in $D$ cannot be empty, otherwise the sets $D_1$ and $D \setminus D_1$ would form a decompositon of $D$ in non-empty disjoint closed subsets. Let $x \in \operatorname{bd} D_1$. Choose a small closed ball $B \subset D$ with center $x$. Then $B \subset D_1$ since $B$ is the disjoint union of the sets $B \cap D_i$ and $B \cap D_1 \ne \emptyset$. We conlude $x \in \operatorname{int} B \subset \operatorname{int} D_1$, hence $x \notin \operatorname{bd} D_1$, a contardiction.


*Let $U_\alpha$ be the components of $\mathbb R^n \setminus C_k$. These are open subsets of $\mathbb  R^n$ and there are at least two of them. By 1. it is impossible that $U_\alpha$ is the union of any subfamily of $\{C_i\}$. Thus each $U_\alpha$ must contain a component of $\mathbb R^n \setminus C$. Therefore $\mathbb R^n \setminus C$ has at least two compenents.
