The Shapovalov radical is the maximal submodule

Contravariance makes $R:=\mathrm{rad}{S}_{\lambda }$ a submodule. It is proper because $S_{\lambda }(v_{\lambda },v_{\lambda })=1$, hence $R\subseteq J(\lambda )$.

Let $N$ be proper. Its stability under $\mathfrak{h}$ lets one project every finite weight decomposition into $N$, so $N$ is the sum of its weight intersections; its $\lambda$-intersection is $0$. If $w\in N$ has weight different from $\lambda$, choose $h$ separating the weights and move $h$ across the form to get $S_{\lambda }(v_{\lambda },w)=0$, hence $S_{\lambda }(v_{\lambda },N)=0$.

For $u\in U(\mathfrak{g})$, $\tau (u)N\subseteq N$, so contravariance gives $S_{\lambda }(u{v}_{\lambda },N)=S_{\lambda }(v_{\lambda },\tau (u)N)=0$. The cyclicity of $v_{\lambda }$ therefore gives $N\subseteq R$. In particular $J(\lambda )\subseteq R$.

Hence $R=J(\lambda )$. Quotienting a bilinear form by its radical is well defined and nondegenerate, which yields the stated form on $L(\lambda )$.