The $k=1$ case

For any nonempty finite simple graph $G$, there is exactly one function $V(G)\to \{1\}$, and it is constant. Consequently every copy in $G$ is monochromatic for that function. By the membership definitions, $G\in {\mathcal{R}}^1(H)$ exactly when $G$ contains an ordinary copy of $H$, and $G\in {\mathcal{R}}_{\mathrm{i}\mathrm{n}\mathrm{d}}^1(H)$ exactly when $G$ contains an induced copy of $H$. In particular, the identity vertex map gives both an ordinary and an induced copy of $H_i$ in itself, proving both self-memberships for each $i$.

By finite mutual-embedding rigidity and non-isomorphism, at least one of the ordinary embeddings $H_1\to H_2$ and $H_2\to H_1$ does not exist. Choose indices $a,b$ with $\{a,b\}=\{1,2\}$ such that $H_a$ has no ordinary embedding into $H_b$. Then $H_b$ contains neither an ordinary nor an induced copy of $H_a$: an induced embedding would also be an ordinary embedding. Thus $H_b$ belongs to both one-colour classes of $H_b$ but to neither corresponding class of $H_a$. The same graph therefore distinguishes each pair of classes, establishing both inequalities.