On a MJE13005A NPN transistor.
So why is the VCBO breakdown voltage always greater than the VCEO?

 

That's true of virtually all bipolar transistors.
When you measure Vceo, the base is open, and the leakage current into the base junction when you get near the breakdown voltage, will start to turn on the transistor due to the transistor current gain, increasing the Ice to the breakdown level.
Measuring Vcbo does not generate this gain function, so the voltage is higher before the breakdown current is reached.

Make sense?