Two-connected subgraphs stay in one incidence-forest piece

Every edge of $R$ belongs to a piece by the union defining $R$. Distinct hyperedges have at most one common vertex: two common vertices and the two hyperedge nodes would form a four-edge cycle in the incidence forest. In a forest there is at most one path between any two nodes: otherwise, after following two distinct paths from one endpoint to their first divergence and then to their first subsequent meeting, the two resulting internally disjoint path segments form a cycle.

The connected graph $H$ has an edge $xy$; choose a hyperedge $e$ whose piece contains it. Suppose that $H$ has a vertex $z$ outside $e$. A path in $H$ from $x$ to $z$ expands to an incidence walk by replacing each graph edge $ab$ with $a,f,b$, where a piece indexed by $f$ contains $ab$. Thus the node of $z$ is connected to the node of $e$. Let $P$ be their unique incidence path. Its first node after the node of $e$ is a vertex node $v$ with $v\in e$. At least one of $x,y$, say $w$, differs from $v$. The path from $w$ to the node of $e$ is the single incidence edge $w,e$, so uniqueness forces the path from $w$ to $z$ to pass through $v$. Consequently no incidence walk from $w$ to $z$ avoids $v$: such a walk would contain a path avoiding $v$, contrary to uniqueness.

Since $H-v$ is connected and contains $w$ and $z$, it has a $w$–$z$ path. Expanding its edges into incidence walks as above gives a walk avoiding the vertex node $v$, a contradiction. Hence every vertex of $H$ lies in $e$. Each edge $ab$ of $H$ belongs to some piece $J_f$; since $a$ and $b$ are distinct members of both $e$ and $f$, step 1.1 gives $f=e$. Thus both the vertex and edge sets of $H$ are contained in those of $J_e$. The final assertion was established in step 1.1.