Union of two sets in tripartition of an open ball contains another open ball Claim. Let $B$ be an open ball in $\mathbb{R}^d$. Let $X,Y,Z\subset\mathbb{R}^d$ be non-empty, open and disjoint such that $B$ is the interior of ${cl}(X\cup Y\cup Z)$. Then, there exists an open ball $B'\subset B$ such that $B'\not\subset X,Y,Z$ and $B'\subset {cl}(X\cup Z)$ or $B'\subset {cl}(Y\cup Z)$.
I cannot come up with a proof for this seemingly obvious claim. Any help is much appreciated.
 A: Suppose $d = 2$. Let $B$ be the open unit disk about the origin.

*

*Let $U_1$ be the union of two disjoint open disks of radius $\frac 12$ inside $B$.

*In each component of $B \setminus \overline U_1$, put an open disk of maximal radius. Let $U_2$ be the union of these disks.

*In each component of $B \setminus \overline {U_1 \cup U_2}$, put an open disk of maximal radius. Let $U_3$ be the union of these disks.

*Etc.

Let $X = \bigcup_k U_{1+3k}, Y = \bigcup_k U_{2+3k}, Z=\bigcup_k U_{3+3k}$. Any neighborhood of a point of $B\setminus X\cup Y\cup Z$ will intersect all three of them.
Connectedness might be enough to avoid this problem in $\Bbb R^2$ (I haven't thought it out, so it may not be enough even there), but in $\Bbb R^3$, you can exploit the other dimension to connect all the pieces of $X, Y, Z$, without breaking the intermingling.
You can also do something similar in $\Bbb R$. You just have to be more artificial about leaving holes to be filled later. But in $\Bbb R$, connectedness is definitely sufficient to prove your result.
