Does Cauchy sequence of disjoint and closed subsets converge to a non-empty set in a Banach space? Suppose, $X$ is a banach space. For any $x,y \in X$, we define $d(x,y) = |x-y|$, For any $A, B \subseteq X$, we define:
$$d(A, B) = \inf_{x \in A, y\in B}{d(x,y)}$$
Say,$(K_n)_{n \in \mathbb{N}}$ is a Cauchy sequence of disjoint and closed subsets of$X$. We say that:
$$K = \{x \in X: \lim_{n \to \infty}d(\{x\}, K_n) = 0\}$$ is the limit of $(K_n)_{n \in \mathbb{N}}$.
Is $K$ necessarily non-empty?
Added: Thank you all for your examples and corrections. My naive attempt is to find a condition stronger than sequential completeness. I hope it will work by adding "disjoint and closed".
 A: Taking my comment, adjusting to make it closed and disjoint.  In the plane, let $A_0, A_1, A_2$ be the three sides of an equilateral triangle, all of length $1$.  They are closed but not disjoint.  Define sequence $K_n$ as follows:  If $n=3m$, let $K_n$ be $A_0$ moved radially outward by distance $1/n$.  If $n=3m+1$, let $K_n$ be $A_1$ moved radially outward by distance $1/n$.  If $n=3m+2$, let $K_n$ be $A_2$ moved radially outward by distance $1/n$.  Then this is a Cauchy sequence, $d(K_i,K_j) \le 1/i+1/j$,
but its "limit" is empty.
A: Let $\Omega \subset \mathbb{R}$ be a countable set. Let $\epsilon >0$. It should be clear that we can choose an increasing sequence $x_1,x_2,...$ such that $x_{k+1}-x_k \in (\frac{\epsilon}{2},\epsilon)$, and $x_k \notin \Omega$ for all $k$. The sequence is, of course, countable, and also  closed since all points are isolated. In addition, we must have $\lim_k x_k = \infty$.
Consequently we can create a collection of sequences $k \to x_k^n$ such that $x_{k+1}^n-x_k^n \in (\frac{1}{2^{n+1}},\frac{1}{2^{n}})$ such that $\{x_k^n\}_k \cap \{x_k^m\}_k = \emptyset$ whenever $n \neq m$.
Now let $K_n = \{x_k^n\}_k \cap [n,\infty)$. Then $K_n$ is closed and the sets $\{K_n\}_n$ are pairwise disjoint. By construction, we have $d(K_n,K_m) \leq \frac{1}{2^{\min(n,m)}}$, so presumably is it 'Cauchy'. However, it should also be clear that $K = \emptyset$. 
