Assume that $M$ is a two dimensional manifold with (pseudo-)Riemannian metric $g$. In some local chart $(\tau,\sigma)$ we have $$ g=A(\tau,\sigma)d\tau^2+B(\tau,\sigma)d\sigma^2+C(\tau,\sigma)(d\tau\otimes d\sigma+d\sigma\otimes d\tau ). $$
I am aware that this metric is always diagonalizable (in fact, all two dimensional metrics are conformally flat), however if we make the additional condition that we must keep the $\sigma$ coordinate fixed, I'm stuck.
It is clear that the metric is diagonized, if we find an exact form $dt$ which is orthogonal to $d\sigma$.
This orthogonality may be expressed in local coordinates as $$ g^{ij}\partial_i t\partial_j \sigma=g^{i \sigma}\partial_i t=0. $$ Renaming $g^{i \sigma}=V^i$, this is just $V^i\partial_i t=0$. The functions $A,B,C$ are arbitrary (as long as they form a metric), so the inverse metric functions $g^{\tau\sigma},g^{\sigma\sigma}$ are essentially arbitrary.
So the problem is equivalent to asking whether an arbitrary smooth vector field (in 2 dimensions) has a (locally) exact annulator.
I can feel this is very easy to prove, but I am unable to. I have no idea how to show that the equation $$ V^\tau(\tau,\sigma)\frac{\partial t}{\partial\tau}+V^\sigma(\tau,\sigma)\frac{\partial t}{\partial \sigma}=0 $$ is soluble.
Any other method is also welcome. I tried to look for situations in which Poincaré's lemma or Frobenius' theorem can be applied to help out - to no avail, but as I said, I think I'm missing something trivial and obvious.