Equivalence of definitions of Killing vector field I read in wikipedia the following two definitions of Killing Vector $X$:

$$\nabla_{\mu}X_{\nu}+\nabla_{\nu}X_{\mu}=0$$
$$ g(\nabla_Y X,Z)+g(Y,\nabla_Z X)=0$$

I have problems deducing the second from the first. My try:
$$\nabla_{\mu}X_{\nu}+\nabla_{\nu}X_{\mu}=0 \\ \iff \\ \nabla_{\mu}g(X,-)(e_{\nu})+\nabla_{\nu}g(X,-)(e_{\mu})=e_{\mu}(g(X,e_{\nu}))-g(X,\nabla_{\mu} e_{\nu})+e_{\nu}(g(X,e_{\mu}))-g(X,\nabla_{\nu}e_{\mu}).$$
But in this last part I don't have the Levi Civita connection acting on X. How can I finish? I'm trying to use the definition of the Levi Civita connection but I get some terms of the form $g([X,Y],Z)$ and I'm not sure what should I do with them.

Making a judicious choice of indices, we can write
$$g(\nabla_Y X, Z) + g(Y, \nabla_Z X) = 0$$
in abstract index notation as
$$g_{ac} \nabla_b X^a Y^b Z^c + g_{ab} \nabla_c X^a Y^b Z^c = 0.$$
Lowering indices using the metric and factoring gives
$$(\nabla_b X_c +\nabla_c X_b) Y^b Z^c = 0 .$$
Since $Y, Z$ are arbitrary, we have
$$\nabla_b X_c +\nabla_c X_b = 0$$
as desired.
Remark The most geometrically natural definition of Killing field, by the way, is probably $$\mathcal L_X g = 0 ,$$ which says that $g$ is (infinitesimally) invariant under the flow of $X$. Using the formula for the Lie derivative in terms of a torsion-free connection quickly leads to the index-free formula.
