Inductive colouring avoiding H1

Applying F6 to $H_2$ and $I$ shows that $F$ is nonempty. The constant assignment $c(v)=1$ on $V(A)$ is a $k$-colouring, even if $V(A)$ were empty, since $k\ge 2$ and colours need not all occur. By the property of $A$ in F5 this colouring contains an induced copy of $F$. Its embedding maps a vertex of $F$ to a vertex of $A$, so $A$ is nonempty. Now $A_1=H_2$ is nonempty because it has an edge, and for every $2\le j\le k$ the core in $A_j$ is nonempty because its vertex set is in bijection with $V(A)$. Thus all $A_j$, in particular $A_{r-1}$ and $A_r$, are nonempty finite graphs.

By step 1.1, $A_{r-1}$ is nonempty, so by nonmembership choose an $(r-1)$-colouring of $A_{r-1}$ with no monochromatic induced copy of $H_1$. For each attachment, transport this colouring through an isomorphism from $A_{r-1}$ onto that part and relabel each colour $i$ as $i+1$. The relabelling is injective and takes values in $\{2,\dots ,r\}$. A monochromatic induced $H_1$ in a transported part would pull back through the inverse isomorphism to one in $A_{r-1}$, and injective relabelling cannot create monochromaticity. Thus every attachment avoids such a copy.

Give every core vertex colour $1$. The core and attachment parts are disjoint and cover $V(A_r)$, so these assignments form an $r$-colouring. The core has no induced $H_1$, since it is isomorphic to $A$; distinct attachments have no edges between them, and colour $1$ is absent from every attachment. Applying the connected-target avoidance fact to this partition and the colourings from step 2.1 shows that $A_r$ has no monochromatic induced copy of $H_1$. Since $A_r$ is nonempty by step 1.1, the definition of the induced Ramsey class therefore gives $A_r\mathrm{\notin }{\mathcal{R}}_{\mathrm{i}\mathrm{n}\mathrm{d}}^r(H_1)$.