Assume that in spacetime ($M,g_{ab}$) there is a hypersurface generated by a set of independent one-parameter transformations acting on one single point, the generators of these transformations being Killing vector fields in $M$.
Questions: Are they Killing vector fields in the hypersurface mentioned above, possessing the induced metric of $g_{ab}$? And how to prove?
Yes. Assume we have $i$ Killing vectors $X_{(i)}$. A metric on the subspace determined by these vectors is $$h_{cd}=g_{ab}\partial_c X_{(i)}^a\partial_d X_{(i)}^b$$ Since $\mathcal{L}_Xg_{ab}=0$ for Killing vector $X$, we can check if this is still satisfied: $$\mathcal{L}_{X_{(i)}}h_{ab}=g_{ab}\mathcal{L}_{X_{(i)}}(X_{(i)}^aX^b_{(i)})$$ And, as mentioned in the comments below, $\mathcal{L}_X X=[X,X]=0$.
I'm not completely sure what OP is really asking but here are some comments to the question (v3):
If $N\subseteq M$ is a $n$-dimensional submanifold inside a $m$-dimensional Riemannian manifold $(M,g)$, $n\leq m$, then $N$ has an induced metric $h\in\Gamma(T^*\!N\otimes T^*\!N)$ from the ambient space $(M,g)$.
If furthermore the tangent space $TN \subseteq TM$ is generated by vector fields $X_{(1)}, \ldots, X_{(n)}$, then the induced metric in a point $p\in N$ is given as $$ h_p (X_{(i)p},X_{(j)p})~=~g_p (X_{(i)p},X_{(j)p}), \qquad 1~\leq ~i,j~ \leq~ n . $$
By definition, the induced metric $h_{ab}$ is given by
$$h_{ab}|_pu^av^b=g_{ab}|_pu^av^b,$$
where $u^a,v^b$ are arbitrary tangent vector fields to the hypersurface, and this expression is valuated at a point $p$ on the hypersurface.
Let $X^a$ be one of the Killing vector fields, and it induces a diffeomorphism $\phi_t$, so $\phi_t^*g_{ab}=g_{ab}$. Therefore,
$$h_{ab}|_pu^av^b=g_{ab}|_pu^av^b\\=(\phi^*_tg_{ab})|_pu^av^b\\=g_{ab}|_{\phi_t(p)}(\phi_{t*}u^a)(\phi_{t*}v^b)\\=h_{ab}|_{\phi_t(p)}(\phi_{t*}u^a)(\phi_{t*}v^b)\\=(\phi_t^*h_{ab})|_pu^av^b$$
In the 3rd line, we used the fact that $\phi_{t*}u^a,\phi_{t*}v^b$ are both tangent vector fields to the hypersurface, as $X^a$ is one of the generators. In the 4th line, we used the definition of the induced metric again. Since $u^a,v^b$ are arbitrary, we conclude that
$$\phi_t^*h_{ab}=h_{ab}.$$
