The supply is only capable of giving 5A if the load put upon it draws that much. That is the load has to have a resistance low enough to pull 5A out of the supply. You c an tell what resistance it has to be by using Ohm's law. For a 3V supply this is:-
3 / 5 = 0.6 ohm
Now a 20mA LED will not pull 20mA, the maximum current you must put through it is 20mA. It is a non linear load, that is the voltage and current relationship is not a straight line. See http://www.thebox.myzen.co.uk/Tutorial/LEDs.html
So you have a resistor to limit the current. When 20mA is flowing through it it will drop 1.2V (or what ever the data sheet says) across it. So if you connect it to a supply any greater that this without a series resistor you will exceed the maximum current and blow it up.
If you short an LED that means you are connecting the two wires of the LED together, this stops any current flowing through it and so it is perfectly protected from any harm. What is not protected however is the thing trying to drive the LED. If this is a resistor then it will take extra current when the LED is shorted out.
Shorting one end of the LED to the supply while the other end is at ground will blow it because you are allowing unrestricted current to flow. Similarly if one end of the LED is connected to the supply shorting the other end to ground will kill it for the same reason. But shorting the two pins together will keep the LED safe.