A question about Killing vector and Riemann curvature tensor

Permit me the use of Latin indices instead of Greek indices and the convention $\nabla_a K_b=K_{b;a} $. So we wish to prove $\newcommand{\Tud}[3]{{#1}^{#2}_{\phantom{#2}{#3}}}$ $$\Tud{K}{a}{;b c} = \Tud{R}{a}{b c d} K^d$$ where $$\Tud{V}{a}{;b c} - \Tud{V}{a}{;c b} = \Tud{R}{a}{d c b} V^d$$ and $$K_{a ; b} + K_{b ; a} = 0$$

Differentiating the last equation, we get $$K_{a ; b c} + K_{b ; a c} = 0$$ so, relabelling and summing, $$K_{a ; b c} + K_{b ; a c} - K_{b ; c a} - K_{c ; b a} + K_{c ; a b} + K_{a ; c b} = 0$$ hence, $$K_{a; b c} + K_{a ; c b} = R_{b d a c} K^d + R_{c d a b} K^d$$ By the interchange symmetry $R_{a b c d} = R_{c d a b}$, and raising indices, we get $$\Tud{K}{a}{;b c} - \Tud{R}{a}{b c d} K^d = -(\Tud{K}{a}{;c b} - \Tud{R}{a}{c b d} K^d)$$

On the other hand, by the first Bianchi identity and antisymmetry, we have $$\Tud{R}{a}{d c b} = \Tud{R}{a}{b c d} + \Tud{R}{a}{c d b}$$ Hence we get $$\Tud{K}{a}{;b c} = \Tud{K}{a}{; c b} + \Tud{R}{a}{d c b} K^d = \Tud{K}{a}{;c b} + \Tud{R}{a}{b c d} K^d + \Tud{R}{a}{c d b} K^d$$ and therefore $$\Tud{K}{a}{;b c} - \Tud{R}{a}{b c d} K^d = \Tud{K}{a}{;c b} - \Tud{R}{a}{c b d} K^d$$ The conclusion follows.

@C.R. This is my 'simpler proof'; I'm pretty sure it's correct, and simpler than Zhen's one as well.

From the first Bianchi identity [Carroll, (3.132)] $$R_{\mu\nu\rho\sigma}+R_{\mu\rho\sigma\nu}+R_{\mu\sigma\nu\rho}=0$$ we have that, for every vector $V^\rho$, $$\nabla_{[\mu}\nabla_\nu V_{\rho]}=\tfrac{1}{6}\big(R_{\rho\alpha\mu\nu}+R_{\mu\alpha\nu\rho}+R_{\rho\mu\nu\alpha}\big)V^\alpha=0,$$ where in the last equation I used the symmetry properties of the indices in the Riemann tensor to reduce it to the Bianchi identity. This is a very useful formula! I'm quite sure about the index placement thanks to the metric compatibility, $\nabla_\mu g_{\nu\rho}=0$ [Carroll, (1.32)], which implies that the metric $g_{\nu\rho}$ commutes with the covariant derivative $\nabla_\mu$. Now, we take the Killing vector $K^\mu$ and we expand the antisymmetrization in the previous equation: $$0=6\nabla_{[\mu}\nabla_\nu K_{\rho]}=\nabla_\mu\nabla_\nu K_\rho + \nabla_\nu \nabla_\rho K_\mu +\nabla_\rho\nabla_\mu K_\nu- \nabla_\nu\nabla_\mu K_\rho-\nabla_\mu\nabla_\rho K_\nu-\nabla_\rho\nabla_\mu K_\nu.$$ Now we use the Killing's equation [Carroll, (3.174)], $\nabla_{(\nu}K_{\rho)}=0$ or $\nabla_\nu K_\rho =-\nabla_\rho K_\nu$, to simplify the previous equation: $$\nabla_\mu\,\nabla_\nu K_\rho=-\nabla_\mu\,\nabla_\rho K_\nu\quad \Rightarrow\quad 0=2\nabla_\mu\nabla_\nu K_\rho+(\nabla_\nu\nabla_\rho-\nabla_\rho\nabla_\nu)K_\mu+\nabla_\rho\nabla_\mu K_\nu-\nabla_\nu\nabla_\mu K_\rho.$$ Again, using Killing's equations $\nabla_\rho\nabla_\mu K_\nu=-\nabla_\rho\nabla_\nu K_\mu$ and $\nabla_\nu\nabla_\mu K_\rho=-\nabla_\nu\nabla_\rho K_\mu$ we get: $$0=2\big(\nabla_\mu\nabla_\nu K_\rho+(\nabla_\nu\nabla_\rho-\nabla_\rho\nabla_\nu)K_\mu\big)=2\big(\nabla_\mu\nabla_\nu K_\rho+R_{\mu\alpha\nu\rho}K^\alpha\big)$$ $$\nabla_\mu\nabla_\nu K_\rho=-R_{\mu\alpha\nu\rho}K^\alpha=R_{\rho\nu\mu\alpha}K^\alpha.$$ Then, we can rise the index $\rho$ multiplying with the metric $g^{\rho\sigma}$ (and using the metric compatibility): $$\nabla_\mu\nabla_\nu K^\sigma=R^\sigma_{\phantom{\sigma}\nu\mu\alpha}K^\alpha.$$ That's all folks! I hope it would be helpful (and correct)!