For a small Vin-Vout difference, is it worth using an LDO vs a buck regulator?
Dropping 5 to 3.3 V at 250 mA will mean having to lose 0.425 Watt in the LDO, you will need a massive heat sink to make that work.
An LDO will never be more efficient than a buck converter, unless you need so little current that the power used by the regulator itself becomes an issue.
I have a mis-designed PCB right now where I tried doing exactly what you are proposing to turn 5 V into 3.3V at 200 mA and even though I have a large'ish copper plane as a heatsink the LDO still reaches 80 deg C in a few seconds.
I'm currently redesigning my power supply to use a MC34063A converter in stead.
LDOs will not be more efficient: (5 V - 3.3 V) * 250 mA = 0.425 W.
Already quite a lot for smallish (SOT-23) LDOs, at least a DPAK is likely necessary. The design (not the efficiency) could be improved with series resistors at the LDO's input to take heat away from the IC and into the resistors, but make sure the voltage drop across the resistors Rser × Imax doesn't get too big for the highest current that is required. At Imax and at the low end of the available input voltage Vin, min, you still need to meet the LDO's minimum input voltage, i.e.
Vout, max + Vdrop, LDO, max ≥ Vin, min - Rser × Imax.
This trick sometimes help if you can't dissipate all the heat within the LDO's own package and want to spread it across more components. Also, the series resistors in front of the LDO sometimes act as a poor man's short circuit protection, given they can handle the full input voltage for a while.
All this is cheap and dirty, so yes: Might well be worth using a buck.
Many have already given you an opinion on the power efficiency, I would just like to bring up some of the reasons I have seen others do this.
Noise immunity. buck/bost regulators, more broadly [SMPS][1], have very poor noise characteristics. They almost guarantee harmonics at the switching frequency. LDOs do not, they create very smooth power.
Simplicity, you are only dropping a small voltage, keep your circuit clean and your components count low.
This noise immunity is normally one of the main reasons I see this. LDOs cannot be beat on this note, you pay power to get clean output power. The specific reason LDOs are so popular is related to the fact that you can use a buck/boost to get your voltage just barely above the operating voltage of your LDO. I have seen this often in 5V circuits, they boost power to 5.5V and then LDO it to the 5V rail. This gives very low noise high quality power while only suffering a 1/11 power loss, still getting about 90% power efficiency off of the LDO.
So, from this perspective, you could always drop the voltage to 4V with a buck and LDO it, but I would just LDO it and make sure you have it connected to a low resistance thermal path so that the heat is easily dissipated.