= Solution
Work on a regular level surface, so $d\beta\ne0$. A nonzero normal <covector> has the local form $n_a=f\nabla_a\beta$. Extend it as a normal to neighbouring level surfaces. Since the <Levi-Civita connection> is torsion-free,
$$
n_{a;b}=f_{;b}\beta_{;a}+f\beta_{;ab},\qquad \beta_{;ab}=\beta_{;ba}.
$$
Antisymmetrizing the product with $n_c=f\beta_{;c}$ over three indices kills the symmetric second-derivative term and the term containing two copies of $d\beta$. Hence
$$
\boxed{n_{[a;b}n_{c]}=0.}
$$
Equivalently $n\wedge dn=0$, the <Frobenius theorem> condition for <hypersurface orthogonality>.
For the <second covariant derivative of a Killing vector>, the curvature sign needs explicit care. Denote the curvature defined on the cover by $C^a{}_{bcd}$, and obtain its other index positions by ordinary raising and lowering with the <metric tensor>. With $D_{abc}=k_{a;bc}$, the <Killing equation> gives $D_{abc}=-D_{bac}$, while the supplied <Ricci identity> gives
$$
D_{abc}-D_{acb}=C_a{}^d{}_{bc}k_d.
$$
Successive use of these two facts gives
$$
\begin{aligned}
D_{abc}&=-D_{bac}=-D_{bca}-C_b{}^d{}_{ac}k_d\\
&=D_{cba}-C_b{}^d{}_{ac}k_d\\
&=D_{cab}+C_c{}^d{}_{ba}k_d-C_b{}^d{}_{ac}k_d\\
&=-D_{acb}+C_c{}^d{}_{ba}k_d-C_b{}^d{}_{ac}k_d.
\end{aligned}
$$
Substitute $D_{acb}=D_{abc}-C_a{}^d{}_{bc}k_d$. The curvature symmetries and <first Bianchi identity> then imply
$$
2D_{abc}=(C_a{}^d{}_{bc}+C_c{}^d{}_{ba}-C_b{}^d{}_{ac})k_d=-2C_{abc}{}^dk_d.
$$
Thus the identity consistent with the cover convention is
$$
\boxed{k_{a;bc}=-C_{abc}{}^d k_d.}
$$
The plus-sign version is valid for the opposite curvature tensor $\widetilde R=-C$. With ordinary metric index raising, the printed plus sign and the cover's curvature convention cannot both be used. This is the <curvature-sign convention in Killing derivative identities>, not a failure of the <Killing equation>.
An explicit sign check uses $ds^2=e^{2x}dt^2-dx^2-dy^2-dz^2$ and the timelike <Killing vector> $k=\partial_t$. Here $k_t=e^{2x}$, $\Gamma^t{}_{tx}=1$ and $\Gamma^x{}_{tt}=e^{2x}$. Direct differentiation gives $k_{t;xx}=e^{2x}$, whereas the cover definition gives $C^t{}_{xxt}=-1$ and $C_{txx}{}^dk_d=-e^{2x}$. This also verifies the corrected sign without relying on a remembered curvature convention.
To prove the <static Killing field is a Ricci eigenvector>, use the permitted alternative of adapted coordinates. <Hypersurface orthogonality> supplies hypersurfaces orthogonal to $k$. Choose spatial coordinates on one such hypersurface and carry them along the flow of the <Killing vector>, using its flow parameter as $t$. The flow preserves both the <metric tensor> and $k$, so it preserves orthogonality to $k$. Consequently the mixed metric components vanish, and the <Killing equation> makes all coefficients independent of $t$:
$$
ds^2=N(\mathbf x)^2dt^2-h_{ij}(\mathbf x)dx^idx^j,\qquad k^a=(1,0,0,0),\qquad k_a=(N^2,0,0,0).
$$
The only nonzero <Christoffel symbols> involving time can be
$$
\Gamma^0{}_{0i}=\partial_i\log N,\qquad \Gamma^i{}_{00}=Nh^{ij}\partial_jN;
$$
in particular $\Gamma^0{}_{ij}=\Gamma^i{}_{0j}=\Gamma^0{}_{00}=0$. In $C_{0i}$, the derivative of the connection trace vanishes because $\Gamma^a{}_{0a}=0$; the other derivative is either a time derivative of a static coefficient or a derivative of $\Gamma^j{}_{0i}=0$. Every connection-product term contains a vanishing coefficient of these same types. Thus $C_{0i}=0$. The one-form $C_{ab}k^b=C_{a0}$ has only a time component, and is proportional to $k_a$. Therefore
$$
\boxed{k_d C^d{}_{[a}k_{b]}=0.}
$$
Changing the overall curvature sign leaves this zero identity unchanged.
Finally a <perfect fluid> in signature $(+---)$ has $T_{ab}=(\rho+p)u_au_b-pg_{ab}$ and $u^au_a=1$. Contract the <Einstein field equations> with $k^b$, and antisymmetrize with $k_a$. Terms proportional to $g_{ab}$ disappear, as does the Ricci term just proved. This leaves
$$
(\rho+p)(k\cdot u)u_{[a}k_{b]}=0.
$$
Two nonzero timelike vectors cannot be orthogonal: in the fluid's rest <local inertial frame>, $k\cdot u=k^0\ne0$. The hypothesis $\rho+p>0$ therefore forces $u_{[a}k_{b]}=0$. Choosing $k$ future-directed gives the normalized <perfect-fluid alignment in a static spacetime> result
$$
\boxed{u^a=\frac{k^a}{\sqrt{k^bk_b}}.}
$$
The enthalpy hypothesis matters: when $\rho+p=0$, the <stress-energy tensor> is proportional to the <metric tensor> and does not select a velocity.
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