Disconnecting a totally disconnected set with a ball. Suppose that $K$ is a compact subset of ${\mathbb R}^2$ that is totally disconnected (meaning that for any two distinct points
$x, y\in K$, there are disjoint open subsets $U,V\subseteq {\mathbb R}^2$, such that $x\in  U$, $y\in  V$ and $K\subseteq U\cup V$).
Does there necessarily
exist $x_0\in {\mathbb R}^2$, and $r>0$, such that $K$ intersects both the interior and the exterior of the ball $B(x_0,r)$, but not
its boundary?
You are welcome to replace the metric in ${\mathbb R}^2$ by one that is equivalent to the usual metric, and hence
change the shape of a ball to something else, such as a regular polygon.
 A: There are planar discontinuums (subsets homeomorphic to the Cantor set) which cannot be separated by any closed convex curve. Below is a sketch of a construction that follows the same idea as the one of the  Antoine's necklace. The latter is a discontinuum in the 3-space $E^3$ which cannot be separated by any topological 2-dimensional sphere in $E^3$.
Start with the "solid half-annulus" $C_1=A_1\subset E^2$. Then inside of $A_1$ inscribe a chain $C_2$ of smaller half-annuli $A_2^i$:
$$
C_2= A_2^1\cup A_2^2 \cup ... \cup A_2^{n_2}.
$$
Repeat this by inscribing a chain $C_3$ inside of each $A_2^i$, etc. See the figure below (sorry for the poor quality).  Now, set
$$
C:= \bigcap_{i=1}^\infty C_i.
$$

The intersection $C$ will be a discontinuum which cannot be separated by any convex curve in $E^2$. The reason for the non-separation property is that every chain $C_i$  cannot be separated by a convex curve. It would be interesting to determine for what range of Hausdorff dimensions of planar discontinua this is possible. It is easy to see that if Hausdorff dimension of a compact subset $K\subset R^2$ is $<1$, then the orthogonal projection of $K$ to a straight line has Hausdorff dimension $<1$. From this it follows that every discontinuum of Hausdorff dimension $<1$ can be disconnected by a convex curve (probably even a round circle). My guess is that for all $s>1$ there exists a planar discontinuum of Hausdorff dimension 1 which cannot be separated by a convex curve. What happens in the case $s=1$, I cannot even guess.
