4 - 20 ma current loop, Short circuit isolator
To address both parts of your question:


What's wrong with the current circuit: 
So remember that a MOSFET is not actually a symmetric device. It can't turn-off current going in both directions, only in one direction. Your schematic has current flowing from source to drain on the MOSFET, but that is the direction of the body-diode in a MOSFET as well. Not all circuit symbols show it, which can lead to this sort of simple mistake, but if you look it at the circuit symbol from the datasheet of the very MOSFET part number you used, you can see the more complete circuit symbol that shows the diode: 




So even after you turn-off the MOSFET in the circuit you have, it will still conduct current through that body-diode. 


Practical note:
This kind of circuit for fault-isolation is generally in the class of ICs known as E-Fuses, and can be bought in a convenient form factor from the likes of TI or STM or Analog Devices. You can even have it re-attempt to turn on in short pulses that won't take down the system but allow you to tell when or if the short has been removed from the system.