Disjoint-copy witness against repetition

Fact F1 gives the existence of $A$ and, for every permitted choice of $A$, gives $A\mathrm{\notin }{\mathcal{R}}_{\mathrm{i}\mathrm{n}\mathrm{d}}^k(H)$. Fix such an $A$ and an arbitrary colouring $c:V(qA)\to \{1,\dots ,k\}$.

For each $i\in \{1,\dots ,q\}$, pull $c$ back along the tagged copy map $v\mapsto (i,v)$ to a $k$-colouring of $A$. Since $A\in {\mathcal{R}}_{\mathrm{i}\mathrm{n}\mathrm{d}}^k(L)$, this colouring gives an induced embedding of $L$ whose image has one colour. Compose it with the tagged copy map; the resulting induced copy of $L$ lies in the $i$th part and is monochromatic for $c$. Choose one such copy and its colour for each of the finitely many indices.

Because $q=k(t-1)+1>k(t-1)$, F5 gives distinct indices $i_1,\dots ,i_t$ whose chosen copies of $L$ have the same colour. For each $j$, restrict the induced embedding of $L$ in the $i_j$th part to the component copy $H_j$ inside $L$. This yields an induced embedding of $H_j$ in that part, and all of its image vertices have the common colour.

Define a map on $V(H)$ by using the embedding selected for $H_j$ on $V(H_j)$. The component vertex sets partition $V(H)$, and the target parts are disjoint, so this map is well-defined and injective. Within a component it preserves and reflects adjacency; between different components there are no edges in $H$ and, because the target parts are distinct, no edges in $qA$. Thus the map is an induced embedding, and its image is monochromatic. As $c$ was arbitrary, $qA\in {\mathcal{R}}_{\mathrm{i}\mathrm{n}\mathrm{d}}^k(H)$.