Complete-graph separator when orders differ

Let $c$ be any $k$-colouring of $V(G)$. Since $\mathrm{\mid }V(G)\mathrm{\mid }=k(n-1)+1>k(n-1)$, a colour class contains at least $n$ vertices. Let $C$ be such a class. Since $C\subseteq V(G)$ and $V(G)$ is finite, F6 makes $C$ finite; put $m=\mathrm{\mid }C\mathrm{\mid }\ge n$. By F5 choose bijections $e:C\to \{0,\dots ,m-1\}$ and $b:V(H_1)\to \{0,\dots ,n-1\}$. The first $n$ indices belong to $\{0,\dots ,m-1\}$, so $f(x)=e^{-1}(b(x))$ defines an injection from $V(H_1)$ into $C$, bijective onto the vertices indexed by those first $n$ indices. Every edge of $H_1$ maps to an edge of $G$, because $G$ is complete. The map therefore gives a monochromatic ordinary copy of $H_1$. As $c$ was arbitrary, $G\in {\mathcal{R}}^k(H_1)$.

Since $k(n-1)+1\le kn$, there is a $k$-colouring of $V(G)$ in which every colour class has at most $n$ vertices. If $J$ were a monochromatic ordinary copy of $H_2$, its isomorphism from $H_2$ would give a bijection $V(H_2)\to V(J)$. By F5, $\mathrm{\mid }V(J)\mathrm{\mid }=\mathrm{\mid }V(H_2)\mathrm{\mid }=\mathrm{\mid }{H}_2\mathrm{\mid }>n$. All vertices of $J$ would lie in one colour class $C$, which is finite by F6 since $C\subseteq V(G)$. Since $V(J)\subseteq C$, F6 gives $\mathrm{\mid }V(J)\mathrm{\mid }\le \mathrm{\mid }C\mathrm{\mid }\le n$, a contradiction. This colouring witnesses $G\mathrm{\notin }{\mathcal{R}}^k(H_2)$. Together with step 1.1, this proves the stated separation.