To prove that the symmetry of a connection is equivalent to the symmetry of the Christofell's symbols in $i$ and $j$, i.e $$\nabla_XY-\nabla_YX-[X,Y]=0\iff \Gamma_{ij}^k=\Gamma_{ji}^k$$ I have thought the following. In coordinates we have $X=\xi^i\partial_i$ and $Y=\eta^j\partial_j$, so:
$$\color{red}{\nabla_XY}-\color{blue}{\nabla_Y X}-\color{green}{[X,Y]}=\color{red}{\xi^i(\eta^j\Gamma_{ij}^k\partial_k+\eta^{j}_{,i}\partial_j)}-\color{blue}{\eta^j(\xi^i\Gamma_{ji}^k\partial_k+\xi^{i}_{,j}\partial_i)}-\color{green}{(\xi^i\eta^j_{,i}-\eta^i\xi^j_{,i})\partial_j}=\xi^i\eta^j\Gamma_{ij}^k\partial_k-\eta^j\xi^i\Gamma_{ji}^k\partial_k=\xi^i\eta^j(\Gamma_{ij}^k-\Gamma_{ji}^k)\partial_k=0\iff \Gamma_{ij}^k=\Gamma_{ji}^k $$ Do you think it is all right? I am not sure that I can write $$\xi^i\eta^j\Gamma_{ij}^k\partial_k-\eta^j\xi^i\Gamma_{ji}^k\partial_k=\xi^i\eta^j(\Gamma_{ij}^k-\Gamma_{ji}^k)\partial_k$$ I think yes since $\xi^i \eta^j$ is a product of functions and so the commutative property holds. Can you confirm me this or disprove? Thanks in advance!
Yes, That is correct. Just plug in carefully $\nabla_XY=(X^i(\partial_iY^k)+X^iY^j\Gamma_{ij}^k)\partial_k$ and note that the components are product of real functions so $X^iY^j=Y^jX^i$ (like $f(x,y,z)\times g(x,y,z)=g(x,y,z)\times f(x,y,z)$).
You can also verify it just for $\partial_i$ and $\partial_j$ for simplicity because $T(X,Y):=\nabla_XY-\nabla_YX-[X,Y]$ is a tensor (known as torsion tensor): $$T_{ij}=\nabla_{\partial_i}\partial_j-\nabla_{\partial_j}\partial_i-[\partial_i,\partial_j]=\nabla_{\partial_i}\partial_j-\nabla_{\partial_j}\partial_i=(\Gamma_{ij}^k-\Gamma_{ji}^k)\partial_k.$$
The RHS is a vector, so it is zero iff its components is. therefore $T=0\stackrel{\text{it is tensor}}{\iff} T_{ij}=0\iff \Gamma_{ij}^k=\Gamma_{ji}^k$.