Given a graph $G = (V,E)$, define a packing in $G$ as a family of connected pairwise-disjoint subsets of $G$. I.e., a family $\mathbf{X} := (X_1,\ldots,X_k)$ such that for each $i,j\in \{1,\ldots,k\}$, $X_i$ is a connected subgraph of $G$, and $X_i$ and $X_j$ are disjoint (in both edges and vertices).
Suppose $G$ is a tree graph, and we have two different packings in $G$, say $\mathbf{X}$ and $\mathbf{Y}$. Then there exists at least one part in at least one of these packings, that intersects at most a single part in the other packing. I.e., there exists some index $i$ such that either $X_i$ intersects at most a single part of $\mathbf{Y}$, or $Y_i$ intersects at most a single part of $\mathbf{X}$.
Proof. Construct a bipartite undirected graph $H$ with the parts of $\mathbf{X}$ on one side and the parts of $\mathbf{Y}$ on the other side, and there is an edge between each two parts whenever they intersect. If the claim is not true, then each vertex in $H$ has degree at least 2, so $H$ has a cycle, say $X_1 - Y_1 - \cdots - X_k - Y_k - X_1$. This cycle corresponds to a cycle in $G$: starting from an edge of $X_1$, there is a path through $Y_1$ to $X_2$, then a path through $Y_2$ to $X_3$ (which must be different from the previous path since $X_1$ and $X_2$ are disjoint), and so on, until we are back at $X_1$. But this cycle contradicts the assumption that $G$ is a tree. Therefore the claim is true.
Does this claim generalize to three or more packings? I.e., given packings $\mathbf{X}^1,\ldots,\mathbf{X}^n$ of the same tree graph $G$, does there always exist a part in one of the packings, that intersects at most a single part in every other packing?
The claim is easy to prove for a path graph: just start at the leftmost end of the path, and find a part whose rightmost endpoint is the leftmost of all parts. This part must intersect at most one part (the leftmost one) in every packing. Can this be generalized to arbitrary tree graphs?
Let’s prove the claim by induction with respect to the number $r$ of vertices of $G$.
If $r=1$ the claim is trivial.
Suppose that the claim is proved for all trees with $r$ vertices, and $G$ is a tree with $r+1$ vertices. Then $G$ has a leaf $v$.
If there exists a given packing whose part is a one-vertex set $\{v\}$, then it is the required part.
Otherwise, each part of each of given packings containing $v$ contains also its unique neighbor $u$. Now remove $v$ from $G$ and each part of each of the given packings. By the inductive hypothesis, there exists a part $P$ in one of the modified packings, that intersects at most a single part in every other residual packing. If $v$ was not removed from $P$, then $P$ is a required part for the given packings; otherwise $P$ contains $u$, so $P$ with an added vertex $v$ and an edge $\{u,v\}$ is a required part for the given packings.