1. Count the valence electrons around xenon

   In $XeO_2F_2$, xenon is the central atom.

   • Xenon has $8$ valence electrons.
   • Each fluorine forms one single bond with Xe.
   • Each oxygen in such oxofluorides is treated as bonded through a double bond in the best Lewis structure.

   So around Xe, we have:

   • $2$ Xe–F single bonds
   • $2$ Xe=O double bonds
   • and then check for lone pair(s).

2. Determine the steric number

   In VSEPR theory, each bond region counts as one electron domain, whether it is a single bond or a double bond.

   Therefore, around Xe:

   • $2$ domains from the two $Xe=O$ bonds
   • $2$ domains from the two $Xe-F$ bonds
   • $1$ lone pair on Xe

   Total electron domains $= 5$.

3. Electronic geometry

   For steric number $5$, the electron-pair geometry is: $\text{trigonal bipyramidal}$

4. Molecular shape

   With $5$ electron domains and $1$ lone pair, the arrangement is of type: $AX_4E$

   The molecular shape for $AX_4E$ in a trigonal bipyramidal arrangement is: $\text{see-saw}$

5. Check the options

   • A: trigonal bipyramidal $\to$ this is the electron-pair geometry, not the molecular shape.
   • B: square planar $\to$ corresponds to $AX_4E_2$, not applicable here.
   • C: tetrahedral $\to$ corresponds to $AX_4$, not applicable here.
   • D: see-saw $\to$ correct for $AX_4E$.

6. Final answer

   $\boxed{\text{see-saw}}$

So, the correct option is D.