On page 257 of Griffiths' Quantum book, he says "Typically, the perturbation will break the degeneracy: As we increase $\lambda$, the common unperturbed energy $E^0$ splits into two. Going the other direction, when we turn off the perturbation, the "upper" state reduces down to one linear combination of $\psi^0_a$ and $\psi^0_b$ [the degenerate states], and the "lower" state reduces to some orthogonal linear combination, but we don't know a priori what these "good" linear combinations will be."
Griffiths provides no justification for the proposition that the perturbation splits degenerate states into degenerate linear combinations, he just states it which isn't very satisfying. It feels right, but I'm not sure how to prove that this must be true. In fact, I now realize that I don't think any justification was given for the stationary states of an unperturbed Hamiltonian being an orthogonal set.
Could you please show me why this must be the case? Thank you!