Is it possible to cover all pairs of points at distance at most 1 by constant number of partitions into sets of diameter at most 1? Let $n$ be a natural number and let $S_n$ be a square $[0,n] \times [0,n]$ in the plane.
We say that a partition $\mathcal{Q} = R_1 \cup \cdots \cup R_t$ of $S_n$ is simple if each of the sets $R_1, \ldots, R_t$ is connected, has positive area and has diameter at most 1.
Question. Does there exist a collection of at most $k$ ($k$ does not depend on $n$) simple partitions $\mathcal{Q}_1, \ldots, \mathcal{Q}_k$ of $S_n$ such that any two points $x,y \in S_n$ of distance at most 1 are contained in at least one of the sets of the partitions, that is there exist $i \in\{1, \ldots, k\}$ and a set $R \in \mathcal{Q}_i$ such that $x,y \in R$?
 A: More generally, replace $S_n$ by $D$, a union of a collection of closed balls of nonzero radius, and suppose $D$ has diameter greater than 1.  Consider partitions of $D$ which are bounded, so each set in a partition has diameter at most 1.  Let us further assume that $D$ contains two open balls with centers $x$ and $y$ of distance exactly 1 from one another.
Consider the unit vector $v$ which is the difference of $x$ and $y$.  Then the points $tv + x$ and $tv+y$ are also distance 1 apart. Let $R_t$ be the partition part which contains these two points.  Then $R_t$ is different from $R_s$ when $t$ and $s$ are distinct real numbers sufficiently small. Thus continuum-many partition parts of diameter 1 are needed to cover just these pairs of points.  In particular, more than countably many parts are needed, so finitely many partitions each with a finite number of parts is not enough.  Thus we can't find a finite set of simple partitions for the poster even in the case of the unit square.
Gerhard "It's About Cardinality, Not Topology" Paseman, 2015.12.02
