Connected subgraphs cannot cross anticomplete parts

Choose $u\in V(H)$ and let $P$ be the unique part containing $u$. For any $v\in V(H)$, connectivity gives a path $u=v_0,v_1,\dots ,v_m=v$ in $H$. Each consecutive pair is an edge of $G$. Starting with $v_0\in P$, each successive vertex must also belong to $P$: otherwise its edge from the preceding vertex would join distinct parts. Thus $v\in P$, including when $m=0$. Consequently $V(H)\subseteq P$.

For an ordinary embedding $f$ of a connected graph $J$ into a disjoint graph family, take the subgraph formed by its image vertices and image edges. It is connected: the image of a path in $J$ is a path in that subgraph, since $f$ is injective and preserves edges. The nonempty tagged parts are pairwise disjoint and still cover the vertex set of the disjoint graph family, so they form a set partition; deleting empty parts changes neither coverage nor the no-crossing-edge property. The image is nonempty because $J$ is connected. Applying step 1.1 to this partition puts $f(V(J))$ in one nonempty tagged part. An induced embedding is also an ordinary embedding, giving the same conclusion.