Why is an energy cost associated with splitting the HOMO spins in singlet oxygen?
Short story: The comment by orthocresol is correct, based on Hund's rules. Given the same spin multiplicity (i.e., the two singlet states), the state with higher orbital angular momentum ($\mathrm{^1\Delta_g}$) wins.

Loosely speaking, one can imagine that comparing the two singlet states, the higher angular momentum implies that the electrons are a bit further out, and suffer less electron-electron repulsion.

Longer story...

The "unified atom" model can apply to simple diatomics - that is, many of the same considerations for the electronic structure can be used for multi-electron atoms. There's discussion in both Mulliken (Rev. Mod. Phys. 1932, 4, 1. and Herzberg's Spectra of Diatomic Molecules).

The vibrational-electronic spectra of $\ce{O2}$ was determined fairly readily:
 (Figure from Chem. Rev., 2003, 103 (5), pp 1685–1758, adapted from Herzberg).

We can assign the lines as discussed by Mulliken in 1932:


  The $\mathrm{^1\Sigma_g^+}$, is 1.62 volts above the $\mathrm{^3\Sigma_g^-}$ according to theoretical considerations,[118] the $\mathrm{^1\Delta_g}$, is probably about halfway between $\mathrm{^1\Sigma_g^+}$, and $\mathrm{^3\Sigma_g^-}$ (cf. Fig 48).


Reference 118 is to Hund and Hückel:


F. Hund, Zeits. f. Physik 63, 726 (1930).
E. Huckel, Zeits. f. Physik 60, 442-3 (1930).


Both Mulliken and Herzberg discuss the Hund's rules in regards to many diatomics - that the triplet will lie lower than the singlet because of higher S, but then the two singlet states will differ based on angular momentum (here L).