Block-hitting deletion lowers maximum block order

Since $J$ has an edge, it has a nontrivial block and $b(J)\ge 2$. If $H$ is edgeless, then $b(H)=1<b(J)$, so assume that $H$ has an edge. Let $B$ be a nontrivial block of $H$ with $\mathrm{\mid }V(B)\mathrm{\mid }=b(H)$.

If $B$ is 2-connected, it is a 2-connected subgraph of $J$, hence lies in a maximal 2-connected subgraph $C$ of $J$. The set $S$ meets $V(C)$, while $V(B)$ is disjoint from $S$. Thus $V(B)$ is a proper subset of the finite set $V(C)$, and $b(H)=\mathrm{\mid }V(B)\mathrm{\mid }<\mathrm{\mid }V(C)\mathrm{\mid }\le b(J)$.

Otherwise $B$ consists of a bridge edge $e=uv$ of $H$ and its endpoints. The edge $e$ is not a bridge of $J$: if it were, its endpoint subgraph would be a nontrivial block of $J$ met by $S$, contrary to both endpoints surviving in $H$. Consequently $u$ and $v$ are connected in $J-e$; otherwise adding $e$ to $J-e$ would merge their distinct components, so deleting $e$ would increase the component count. A path from $u$ to $v$ in $J-e$, together with $e$, forms a simple cycle with at least three vertices. This cycle is 2-connected, since deleting any vertex leaves a connected path. By [F1] it lies in a maximal 2-connected subgraph $C$ of $J$. Therefore $b(H)=2<\mathrm{\mid }V(C)\mathrm{\mid }\le b(J)$.

The edgeless case and the two possible types of $B$ cover all cases, so $b(J-S)=b(H)<b(J)$. No step used independence of $S$.