 Solution

On the extension in part (i), the hypersurface $\mathcal{N}$ has nonzero normal covector $n_a={\partial }_a\lambda$. Its squared norm isThis vanishes on $\mathcal{N}$, which is therefore a null hypersurface. Raising the index givesSince ${\partial }_t={\partial }_v$ at fixed $\lambda$, the original Killing vector field becomes the smooth nonzero field $K={\partial }_v$. The transformed metric tensor is independent of $v$, so $K$ remains a Killing vector field, andThus $K$ is both tangent and normal to $\mathcal{N}$. This is exactly a Killing horizon:The Killing vector field itself does not vanish there; a vanishing squared norm means nullness, not a zero vector.