I learn GR now, and there is a strange thing that I discovered.
It is well-known, that the condition $\nabla_{\mu}g_{\alpha \beta}=0$ is specified, when we choose specific metric-compatible Levi-Civita connection on our manifold.
Otherwise, there is a proof, which does not use properties of Levi-Civita connection at all, but only uses the fact, that covariant derivative has Leibniz rule and respect index raising rules:
$\nabla_{\alpha}A_{\mu}=\nabla_{\alpha}(g_{\mu\nu}A^{\nu})=g_{\mu\nu}\nabla_{\alpha}A^{\nu}+A^{\nu}\nabla_{\alpha}g_{\mu \nu}$
Now we say, that for any vector $A$ the object $\nabla_{\alpha}A_{\mu}$ must be a tensor, namely $\nabla_{\alpha}A_{\mu} \equiv D_{\alpha \mu}$, and it must respect index raising rules, so by definition:
$g^{\mu \nu}\nabla_{\alpha}A_{\mu}=g^{\mu\nu}D_{\alpha\nu}=D_{\alpha}^{\mu}=\nabla_{\alpha}A^{\mu}\implies\nabla_{\alpha}A_{\mu}=g_{\mu\nu}\nabla_{\alpha}A^{\nu}$
Now we immediatly conclude, that in first equation $A^{\nu}\nabla_{\alpha}g_{\mu \nu}=0$, and hence $\nabla_{\alpha}g_{\mu \nu}=0$.
There must be some mistake, but I can't see it.