Diagonalization of the Ricci curvature tensor in dimension 3

It seems to be well-known that in dimension 3, we can simultaneously diagonalize the metric and the Ricci tensor at any given point (see https://mathoverflow.net/questions/80452/diagonalizability-of-the-curvature-operator).

I am unable to prove this myself, and so I am hoping for a proof of this fact or a source which provides the proof. I think the link above has a proof by counting dimensions, but I am looking for a more straightforward proof if there is one.

This is not a particular statement about the dimension 3, but a general statement. As a symmetric bilinear form, the Ricci curvature admits an orthonormal basis of eigenvectors, which trivially diagonalizes the metric as well. See, e,g, this MSE post.