Are all Killing Fields Geodesic Vectofields?

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I will prove here that all Killing vectorfields on a Riemannian manifold $(M,g)$ equipped with Levi-Civita-Connection $\nabla$ are geodesic vectorfields. I know that this is not true, but i don't see where my proof goes wrong.

Let $K$ be a Killing field. Then we have by the Killing equation:

$g(K, \nabla_K K)=-g(\nabla_K K, K)=0$.

From the metricity of LC-connection we get:

$0=Kg(K,\nabla_K K) = g(\nabla_K K, \nabla_K K)+g(K,\nabla_K \nabla_K K) $

thus:

$ g(\nabla_K K, \nabla_K K)=-g(K,\nabla_K \nabla_K K) $

Now i pick a local chart $\phi$ s.t. the Killing field agrees with the first coordinate vectorfield:

$K=\frac{\partial}{\partial x_1}$

(i think for a smooth vectorfield this is always possible).

For Killing vectorfields it is known the following identity with the Riemann tensor:

$(\nabla_a \nabla_b K)_c = R_{dacb} K^d $

thus in the chart $\phi$ we have

$(\nabla_K \nabla_K K)_c= (\nabla_1 \nabla_1K)_c=R_{d1c1} K^d = R_{11c1}=0 $.

Thus we finally we obtain:

$ g(\nabla_K K, \nabla_K K) = K^c (\nabla_K \nabla_K K)_c = R_{1111}=0$

which implies

$\nabla_K K=0$ and therefore $K$ is a geodesic Vectorfield.

Where does this proof go wrong?