Finite outside roots give a finite complementary parabolic

Each element of the finite set $D$ has finite support. Let $F$ be the union of those supports; it is finite, and $\mathrm{span}D\subseteq V_F$. By [F1], $V_K\subseteq V_F$, hence $K\subseteq F$ by linear independence of the simple-root basis. Thus $K$ and $J$ are finite.

Since $J\subseteq I$, [F1] gives a map $\pi :W_J\to \mathrm{Sym}(D)$ defined by $\pi (g)(\beta )=g\beta$ for $\beta \in D$, using [F4]. Each such map is a bijection with inverse $\pi (g^{-1})$, and the geometric action gives $\pi (gh)=\pi (g)\circ \pi (h)$ and $\pi (1)={\mathrm{i}\mathrm{d}}_D$. If $\pi (g)={\mathrm{i}\mathrm{d}}_D$, then $g$ fixes $\mathrm{span}D$, hence fixes $V_K$ and in particular $V_J$ by [F1]. By [F2] applied to finite $J$, the group $W_J$ is the Coxeter group on the restricted matrix. Its restriction to $V_J$ is exactly the canonical representation, since the generator formulas coincide. Faithfulness [F3] gives $g=1$. If $\pi (g)=\pi (h)$, the composition identities give $\pi (h^{-1}g)={\mathrm{i}\mathrm{d}}_D$, so $g=h$; thus $\pi$ is injective. The set $\mathrm{Sym}(D)$ is finite: every map $D\to D$ is determined by one image in the finite set $D$ for each of its finitely many elements, so there are finitely many such maps and hence finitely many bijections. Therefore $W_J$ is finite. For $J=\mathrm{\varnothing }$ the subgroup is the identity and the same conclusion holds.