I'm trying to show the following. Suppose $(M, g)$ is a $2$-dimensional Riemannian manifold with connection $\nabla$. Suppose also that $\nabla$ is metric compatible, and that length extremizing curves $\gamma$ for $g$ are geodesics in the sense that
$$\nabla_{\gamma '}\gamma ' \propto \gamma '$$
Then $\nabla$ is torsion-free. I've got stuck partway through the proof - does anyone know how to finish it off correctly? My working is below.
Let $p$ be an arbitrary point in $M$ and choose a vector field $T$ around $p$ such that $\{T, \nabla_T T\}$ forms a local frame for $M$. In other words
$$ \nabla_T T \not\propto T$$
We now immediately know that the integral curve $\gamma$ of $T$ through $p$ is not length extremizing. Thus there exists a variation $\tilde{\gamma}(t,s)$ of $\gamma$ and a vector field $N = \tilde{\gamma}_*(\partial/\partial s)$ such that
$$L'(s) = \int_{-\epsilon}^\epsilon N_{\tilde{\gamma}(t,s)}|T_{\tilde{\gamma}(t,s)}|dt$$
is nonzero at $s = 0$. Writing this out more explicitly we get
\begin{align} L'(s) &= \int |T|^{-1} g(\nabla_N T, T)dt\\ &= \int |T|^{-1} g (\nabla_T N,T) + Tors(N,T) dt \end{align}
But here I'm stuck. I have no idea how to use this information to prove that the torsion must be $0$. I think perhaps some kind of contradiction argument might work, but I can't quite see how to implement it. Any ideas?