Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is(are)

Molecular orbital (MO) theory gives a direct relation between electronic configuration and measurable properties such as bond order, bond energy and bond length.
Bond order $$= \dfrac{n_b - n_a}{2}$$ where $$n_b$$ and $$n_a$$ are the numbers of electrons in bonding and antibonding MOs respectively.

Option A Ne$$_2$$

Each Ne atom has eight valence electrons (2s$$^2$$ 2p$$^6$$); Ne$$_2$$ therefore has 16 valence electrons.

For elements with atomic number $$\gt 7$$ (O, F, Ne) the MO order is
$$\sigma_{2s}\lt\sigma^{*}_{2s}\lt\sigma_{2p_z}\lt\pi_{2p_x}=\pi_{2p_y}\lt \pi^{*}_{2p_x}=\pi^{*}_{2p_y}\lt\sigma^{*}_{2p_z}$$

Filling 16 electrons gives
bonding electrons $$n_b = 8$$ (σ$$_{2s}$$, σ$$_{2p_z}$$, π$$_{2p_x}$$, π$$_{2p_y}$$)
antibonding electrons $$n_a = 8$$ (σ$$_{2s}^{*}$$, π$$_{2p_x}^{*}$$, π$$_{2p_y}^{*}$$, σ$$_{2p_z}^{*}$$)

Hence $$\text{Bond order} = \dfrac{8-8}{2}=0$$; Ne$$_2$$ is not a stable molecule. Statement A is therefore correct (so it is NOT an incorrect statement).

Option B F$$_2$$

Each F atom contributes seven valence electrons; F$$_2$$ has 14 valence electrons. Using the same MO order as above:

σ$$_{2s}$$ (2) → σ$$_{2s}^{*}$$ (2) → σ$$_{2p_z}$$ (2) → π$$_{2p_x}$$ (2) & π$$_{2p_y}$$ (2) → π$$_{2p_x}^{*}$$ (2) & π$$_{2p_y}^{*}$$ (2)

The highest-energy filled level (HOMO) is the degenerate pair $$\pi^{*}_{2p_x},\;\pi^{*}_{2p_y}$$—both are $$\pi$$-type antibonding orbitals, not σ-type. Hence statement B is incorrect.

Option C Comparison of O$$_2$$ and O$$_2^{+}$$

O$$_2$$ has 12 valence electrons. Configuration: σ$$_{2s}^{2}$$ σ$$_{2s}^{*2}$$ σ$$_{2p_z}^{2}$$ π$$_{2p_x}^{2}$$ π$$_{2p_y}^{2}$$ π$$_{2p_x}^{*1}$$ π$$_{2p_y}^{*1}$$

For O$$_2$$ $$n_b=8,\; n_a=4 \Longrightarrow \text{BO}=\dfrac{8-4}{2}=2$$

O$$_2^{+}$$ has one electron fewer; the electron is removed from a π$$_{2p}^{*}$$ antibonding orbital:

O$$_2^{+}$$ π$$_{2p_x}^{*1}$$, π$$_{2p_y}^{*0}$$

Now $$n_b=8,\; n_a=3 \Longrightarrow \text{BO}=\dfrac{8-3}{2}=2.5$$

Greater bond order implies stronger bond (larger bond energy) and shorter bond length. Therefore O$$_2^{+}$$ has a higher (not smaller) bond energy than O$$_2$$. Statement C is incorrect.

Option D Li$$_2$$ vs. B$$_2$$

Li$$_2$$ (two valence electrons) σ$$_{2s}^{2}$$ ⇒ $$n_b=2,\; n_a=0$$, BO = 1

B$$_2$$ (six valence electrons) σ$$_{2s}^{2}$$ σ$$_{2s}^{*2}$$ π$$_{2p_x}^{1}$$ π$$_{2p_y}^{1}$$ ⇒ $$n_b=4,\; n_a=2$$, BO = 1

Though both have the same bond order, the π(2p)-π(2p) overlap in B$$_2$$ is stronger than the σ(2s)-σ(2s) overlap in Li$$_2$$, producing a shorter bond length for B$$_2$$. Thus Li$$_2$$ indeed has the larger bond length. Statement D is correct (so it is NOT an incorrect statement).

Hence the INCORRECT statements are:
Option B (HOMO of F$$_2$$ is σ-type), Option C (Bond energy of O$$_2^{+}$$ is smaller than that of O$$_2$$).

Option B, Option C