We prove the first item by induction on $n=|E(G)|$ the number of edges in $G$.
$n=0$: trivial.
$n=1$: $E(G)=\{(u,v)\}$. Let $K\in\mathfrak{R}(G,H)$. Two cases:
either $\forall (x,y)\in E(H)$ $\{((u,x)(u,y)), ((v,x)(v,y))\} \subset E(K)$ then $K$ contain a copy of $H$.
or, $\exists (x,y)\in E(H)$ $\{((u,x)(v,x)), ((u,y)(v,y))\} \subset E(K)$ then $K$ contain a copy of $H$.
$n\to n+1$: Suppose the property true for all graphs such that $|E(G)|=n$. Let $G$ be a graph with $|E(G)|=n+1$, let $K\in\mathfrak{R}(G,H)$. Let $E'$ be such that $E(G)=E'(G)\cup\{(u',v')\}$ and $|E'(G)|=n$, let $G'=(V(G),E'(G))$. We define $K'$ as the sub graph of $K$ where we removed all the edges $((u',x),(v',x))$. It's easy to see that $K'\in\mathfrak{R}(G',H)$.
By induction hypothesis either $K'$ contains a copy of $H$, and since $K'$ is contained in $K$, $K$ also contains a copy of $H$. Or, $K'$ contains a copy of $G'$. Let $x'\in V(G)$ be the element shared by all the nodes of this copy. Two cases:
- either, $((u',x'),(v',x,))\in V(K)$ then there is a copy of $G$ in $K$.
- or, $((u',x'),(v',x,))\notin V(K)$. By definition of $K$ that mean $\forall (x,y)\in E(H), \{((u',x),(u',y)),((v',x),(v',y))\}\subset E(K)$ that means there is a copy of $H$ in $K$.
Hence in all cases there is either a copy of $H$ or a copy of $G$ in $K$.
For the second item, by using the first item we know that $\chi(K)\geq\chi(H)$ or $\chi(K)\geq\chi(G)$ hence $\chi(K)\geq k$.