= Solution
On the extension in part (i), the <hypersurface> $\mathcal N$ has nonzero normal <covector> $n_a=\partial_a\lambda$. Its squared <norm> is
$$
g^{ab}n_an_b=g^{\lambda\lambda}=\frac{\lambda^2}{M^2}.
$$
This vanishes on $\mathcal N$, which is therefore a <null hypersurface>. Raising the index gives
$$
n^a=-\frac1M(\partial_v)^a+\frac{\lambda^2}{M^2}(\partial_\lambda)^a.
$$
Since $\partial_t=\partial_v$ 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>, and
$$
\left.n^a\right|_{\mathcal N}=-\frac1M K^a,\qquad
\left.g(K,K)\right|_{\mathcal N}=0.
$$
Thus $K$ is both tangent and normal to $\mathcal N$. This is exactly a <Killing horizon>:
$$
\boxed{\mathcal N=\{\lambda=0\}\text{ is a Killing horizon of }\partial_t=\partial_v.}
$$
The <Killing vector field> itself does not vanish there; a vanishing squared <norm> means nullness, not a zero vector.
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