Prove that the covariant derivative commutes with musical isomorphisms

Suppose I have a covector field $$\omega$$ and a covariant derivative $$\nabla_{X}$$ for some vector field $$X$$ on a Riemannian manifold $$(M, g)$$.

Define $$X^{\flat} \in \mathfrak{X}^{*}(M)$$ as $$X^{\flat}(Y) = g(X, Y)$$, and $$\omega^{\sharp}$$ as the unique vector for which $$\omega(X) = g(\omega^{\sharp}, X)$$; it's easily checked that these musical maps are mutually inverse, so they're isomorphisms between $$\mathfrak{X}(M)$$ and $$\mathfrak{X}^{*}(M)$$.

Anyway, I want to prove that, for the Levi-Civita connection $$\nabla$$ on $$(M, g)$$, we have $$\nabla_{X}(\omega^{\sharp}) = (\nabla_{X}\omega)^{\sharp}.$$ In several books I've been reading, it says that this follows immediately from $$\nabla g = 0$$.

However, I keep getting that this is wrong, for example if $$(E_{1},...,E_{n})$$ is a local orthonormal frame on $$M$$, I get that $$\nabla_{E_{i}}(E_{j}^{*}{^{\sharp}})$$ does not equal $$(\nabla_{E_{i}}E_{j}^{*})^{\sharp}$$. Here's what I've been trying:

1. Finding $$\nabla_{E_{i}}(E_{j}^{*}{^{\sharp}})$$: I start by finding the components of $$E_{j}^{*}{^{\sharp}}$$

$$(E_{j}^{*}{^{\sharp}})^{k} = \sum_{l} g^{kl} (E_{j}^{*})_{l} = \delta_{kj}.$$

Therefore, $$E_{j}^{*}{^{\sharp}} = E_{j}$$. Now, I want to find $$\nabla_{E_{i}} E_{j}^{*}{^{\sharp}} = \nabla_{E_{i}} E_{j}$$.

$$\nabla_{E_{i}} E_{j} = \sum_{k} \Gamma_{ij}^{k}E_{k},$$

where $$\Gamma_{ij}^{k}$$ are the Christoffel symbols with respect to this frame.

2. Finding $$(\nabla_{E_{i}}E_{j}^{*})^{\sharp}$$: again, I begin by finding the components, this time of $$(\nabla_{E_{i}}E_{j}^{*})^{\sharp}$$:

$$((\nabla_{E_{i}}E_{j}^{*})^{\sharp})^{k} = \sum_{l} g^{kl} (\nabla_{E_{i}}E_{j}^{*})_{l},$$ so I want to find $$\nabla_{E_{i}}E_{j}^{*}(E_{l})$$. Since $$\nabla_{E_{i}}$$ is a tensor derivative, we have:

$$0 = \nabla_{E_{i}}(E_{j}^{*}(E_{l})) = \nabla_{E_{i}}E_{j}^{*}(E_{l}) + E_{j}^{*}(\nabla_{E_{i}}E_{l}),$$

so $$\nabla_{E_{i}}E_{j}^{*}(E_{l}) = - E_{j}^{*}(\sum_{k}\Gamma_{il}^{k}E_{k}) = - \Gamma_{il}^{j}$$. Finally,

$$((\nabla_{E_{i}}E_{j}^{*})^{\sharp})^{k} = \sum_{l} g^{kl} (\nabla_{E_{i}}E_{j}^{*})_{l} = -\Gamma_{ik}^{j},$$

so $$(\nabla_{E_{i}}E_{j}^{*})^{\sharp} = \sum_{k} (-\Gamma_{ik}^{j}E_{k}) = -\sum_{k}\Gamma_{ik}^{j}E_{k}$$.

Since I don't know that $$\Gamma_{ik}^{j} + \Gamma_{ij}^{k} = 0$$, I don't see why these two are equal. Did I make a mistake somewhere; is there an easier proof of this fact?

We have that $$(\nabla_X \omega)^\sharp$$ is such that $$g((\nabla_X \omega)^\sharp, Y) = \nabla_X \omega(Y)$$ but, $$\nabla_X \omega(Y) = X\omega(Y) - \omega(\nabla_X Y) = Xg(\omega^\sharp,Y) - g(\omega^\sharp,\nabla_X Y)$$ $$\nabla_X \omega(Y) = g(\nabla_X(\omega^\sharp),Y) + g(\omega^\sharp,\nabla_X Y) - g(\omega^\sharp,\nabla_X Y)$$ Thus $$g((\nabla_X \omega)^\sharp, Y) = g(\nabla_X(\omega^\sharp),Y)$$ for all $$Y$$vector fields in $$M$$. Therefore $$(\nabla_X \omega)^\sharp = \nabla_X(\omega^\sharp)$$.