This problem is somewhat related to some homework I had recently. However, as stated, I don't know if a solution yet exists. I asked some friends and some of my professors, but none of them know how to make significant progress on this problem either, so I figured I'd come here. Any help is appreciated.
The Problem: Let $\mathcal{U}$ be a minimal finite covering of the closed unit disk in $\mathbb{R}^2$ by open disks. ("Minimal" in the sense that if we removed a ball from $\mathcal{U}$, then it would no longer be a covering.) Prove or disprove that we can index $\mathcal{U}$ as $\mathcal{U}=\{U_i\}_{i\in\{1,2,\ldots n\}}$ so that $U_i\cap U_{i+1}\neq\varnothing$ for all $i\in\{1,2,\ldots n-1\}$.
As an example, see the covering below and the arrows which indicate a way to list them in the desired fashion:
Some observations: This isn't true for minimal finite coverings of all sets, or even of convex sets. For example, a triangle can be covered by four open balls so that they can't be listed as desired:
Given a $\mathcal{U}$, we can construct a graph $G$ from it by letting disks in the covering correspond to vertices in $G$ and saying that two vertices are adjacent if their corresponding disks intersect. We call $G$ the graph induced by $\mathcal{U}$. In this form, the given problem is equivalent to finding a Hamilton path in the graph induced by the covering.
It's worth noting that the graph induced by $\mathcal{U}$ might not be planar. In fact, given any $n\in\mathbb{N}$, we can construct coverings whose induced graphs have $K_n$ as a subgraph.
If $\mathcal{U}$ has at least 3 disks, its induced graph is non-separable. Furthermore, every vertex in its induced graph is included in some 3-cycle.
In the picture below, the unit disk is covered by red disks and blue disks, and each blue disk is adjacent only to a red disk. So along a Hamiltonian path each blue disk is followed by a red disk. But this is impossible because there are nine blue disks but only seven red disks.