Step 1: Instead of building the full molecular orbital diagram from scratch, use the isoelectronic trick. Two species are isoelectronic if they have the same total number of electrons, and isoelectronic species have the same molecular orbital filling pattern and hence the same bond order.
Step 2: Count electrons in $O_2^{2-}$: neutral $O_2$ has $8 + 8 = 16$ electrons, and the $2-$ charge adds 2 more, giving $16 + 2 = 18$ electrons.
Step 3: Now think of a well known neutral diatomic molecule with exactly 18 electrons. Fluorine gas, $F_2$, has $9 + 9 = 18$ electrons, the same count. So $O_2^{2-}$ and $F_2$ are isoelectronic and must have the same bond order.
Step 4: The bond order of $F_2$ is well established from its molecular orbital diagram (it has a single bond, F-F), which is 1. Since $O_2^{2-}$ has the same number of electrons filling the same set of molecular orbitals in the same order, its bond order must also be 1.
\[\boxed{\text{Bond order} = 1}\]