Connectedness of balls in a compact, connected metric space Let $(X,d)$ be a compact, connected metric space. For every $\epsilon>0$ define an equivalence relation on $X$ by $x\sim_{\epsilon}y$ if and only if there exists a finite sequence $(x=x_0,x_1,\dots,x_n=y)$ such that $d(x_i,x_{i+1})<\epsilon$. 
Note that the space is connected if and only if for every $\epsilon>0$, the $\epsilon$-equivalence class of every point is the whole space. See this answer. My interest in this collection of equivalence relations is their properties when one restrict them to certain subsets of the space: For a set $A \subset X$ and $\epsilon>0$, define $a \sim_{\epsilon}^{A} b$ if and only if there exists a $\epsilon$-step sequence between $a$ and $b$ contained in $A$.
The following property is something intuitive one could expect to hold in any compact, connected metric space:

Let $U$ and $V$ be open disjoint subsets of $X$ and denote $K:=(U \cup V)^\complement$. Let $\epsilon>0$ and let $u \in U$ and $v \in V$ with $u\sim_{\epsilon}^{U \cup V} v$ through a finite sequence $S_{\epsilon}(u,v) = (u=x_0,x_1,\dots,x_n=v) \subset U \cup V$. Then there exists some $w \in S_{\epsilon}(u,v)$ with $d(w,K)<\epsilon$.

The proof I managed to find consists on the additional assumption that every ball is a connected subset of the space. Assuming this, one can easily see that there exists some ball $B$ of radius $\epsilon$ intersecting $U$ and $V$, so assuming the ball is connected the proof is almost immediate.
I couldn't find a counterexample to this property for compact, connected metric spaces that contain some disconnected ball. Yet, I couldn't prove it when I removed the assumption.
 A: Consider the unit circle $S^1$ with its standard angular distance function $d$. For non-antipodal points $p, q\in S^1$ let 
$pq$ the closed arc with the end-points $p, q\in S^1$ and of the length $<\pi$. Now, for $\epsilon$ satisfying $0<\epsilon<\pi/2$, let $uv, ab$ be two such arcs of the lengths $\epsilon/2$ and $\epsilon/4$ respectively such that $ab$ is contained in the interior of $uv$. Let $X$ denote the metric space obtained by removing the arc $ab$ from $S^1$ (and keeping the distance function). Then $X$ is clearly connected. Let $k\in S^1$ be the point antipodal to the midpoint of the arc $ab$. Let $U, V$ denote the connected components of $X -\{k\}$ containing $u, v$ respectively. Thus, $K=\{k\}$ is the complement to $U\cup V$ in $X$ and $U\cap V=\emptyset$, while $U, V$ are both open in $X$. At the same time, $d(u,k)\ge \pi- \epsilon > \epsilon$ and the same for $v$. Thus, $K$ contains no points within distance $\epsilon$ from the $\epsilon$-chain $\{u, v\}$. 
I strongly suspect that you have in mind a different question.  
