Given a basis of $T_pM (e_i)$. Extend this base to a local orthonormal frame $(E_i)$ with $\nabla E_i (p) = 0 $ I have been working this problem almost 3 days, I would appreciate any help or idea:
Let (M,h) be a Riemannian manifold. For every $ p \in M $ and $ (e_1 , ... , e_n)$ basis of $T_pM$. There exists an orthonormal frame in an neighborhood of p $(E_1, ... , E_n)$ , with $E_i (p) = e_i$ and $\nabla E_i (p) = 0 $
Hint: Fix an orthonormal frame $(\overline E_i)$ near p with $\overline E_i (p) = e_i$ and define $E_i = \alpha_i^j\overline E_i $ with $(\alpha_i^j(x))_{ij} \in SO(n)$ and $\alpha_i^j(p)= \delta_i^j$.
What I have got:
The construction of an orthonormal frame $(\overline E_i)$ near p with $\overline E_i (p) = e_i$. Follows from Gram-Schmidt process with no problem.
Defining $E_i= \sum_j \alpha_i^j\overline E_i$ the frame is still orthonormal, that follows from a direct calculation and the fact that  $(\alpha_i^j(x))_{ij} \in SO(n)$ and $\overline E_i (p) = e_i$ because $\alpha_i^j(p)= \delta_i^j$.
Since $h(E_i,E_j)= \delta_i^j$ then: $h(\nabla_X E_i, E_j) + h(E_i,\nabla_X E_j) = 0 $.
So all I have to see is that: 
$h(\nabla_X E_i, E_j) = h(E_i,\nabla_X E_j) $ 
Writing the above equation with the koszul formula and doing the calculations, Using the antisymmetry of Lie bracket and the fact that the inner product is commutative:
$h(\nabla_X E_i, E_j) - h(E_i,\nabla_X E_j) = -h([E_i,E_j],X) $ 
So all at have to see is that in p: $[E_i,E_j](p) = 0$
That the Lie bracket its 0 at p for the basis constructed in this way! I have been trying to justify this from lie bracket proprieties with no luck

Hint: The actual task is to show that you can choose the functions $\alpha^i_j$ in such a way that the conditions are satisfied, which mainly amounts to fixing the derivatives at $x$ without destroying the fact that the matrix $(\alpha^i_j)$ is orthogonal.
My preferred way to solve the problem would not be by choosing an orthonormal frame and then adapting via the $\alpha^i_j$ but rather use parallel transport to directly construct an appropriate frame.  
