Am I misunderstanding the conditions of the problem or is this a counterexample?

Later: I think the answer to your updated question is yes -- or rather, no, any subgraph induced by disjoint convex sets has no cycles. (However, that would make it a directed acyclic graph or DAG, and not a tree, unless that's computer science terminology.) I have a sketch of what I think should be a valid proof, but filling in the details is a little ugly. Unfortunately, I think there is no completely non-ugly proof, for two reasons:
The power of convexity is a little lost because separating hyperplanes are of no help here. If $A \rightarrow B \rightarrow C$, there is not necessarily a hyperplane separating $C$ from $A \cup B$.
The proposition is not valid in more than 2 dimensions, so any proof must fundamentally depend on the 2D nature of the problem.
That said, here's the fairly ugly proof I got. Suppose for the sake of contradiction that there is a cycle of disjoint compact convex sets, $A_1 \rightarrow A_2 \rightarrow \cdots \rightarrow A_n \rightarrow A_1$. Each set $A_i$ contains a pair of points $p_i$ and $q_i$ such that $q_{i-1} \rightarrow p_i$ and $q_i \rightarrow p_{i+1}$ (where $i+1$ and $i-1$ are modulo $n$, of course). Therefore, it suffices to prove the impossibility of a cycle on the line segments $p_1 q_1 \rightarrow p_2 q_2 \rightarrow \cdots \rightarrow p_n q_n \rightarrow p_1 q_1$.
Consider the rightmost "fringe" of the first $i$ segments, as a function of $y$; that is, let $f_{i}(y) = \sup \\{x : (x,y) \in p_1 q_1 \cup p_2 q_2 \cup \cdots \cup p_i q_i\\}$. While $f_i$ is not continuous, I think you can prove by induction that if $p_1 q_1 \rightarrow p_2 q_2 \rightarrow \cdots \rightarrow p_i q_i \rightarrow p_{i+1} q_{i+1}$, then the discontinuities of $f_i$ are not "visible" to $p_{i+1}$. What I mean is that $f_i$ jumps leftwards rather than rightwards as you move in $y$ away from $p_{i+1}$, so there is no way for $q_{i+1}$ to sneak in to the left of $f_i$ without the segment $p_{i+1}q_{i+1}$ intersecting $f_i$ somewhere. Therefore, every $p_{i+1}$ and $q_{i+1}$ is to the right of $f_i$. In particular, $q_n$ is to the right of $f_{n-1}$. But $p_1$ is on or to the left of $f_{n-1}$, so $q_n \rightarrow p_1$ is impossible.