MOSFET common source amplifier output impedance calculation
You simply forget that you are dealing here with a small-signal equivalent model (AC signal only).

So, in this case, shorting the transistor gate to ground means that any AC signal present at the gate is shorted to ground. Hence, no AC voltage is present at the transistor gate. But the DC condition (quiescent currents and voltages ) should stay unchanged.

This is why we also "short-out" the DC voltage source in a small-signal analysis.  

The ideal DC voltage source has 0Ω internal resistance. And that's why AC-signals are shorted by DC voltage source. DC voltage is always constant so for any change in current there is no change in the voltage. So there is 0 Ohm internal resistance. Additional in real life circuit we always use a bypass capacitor connect parallel to DC Voltage. And this capacitor will short all AC-signal to the ground.

For example, if we have a 9V DC voltage source, now if we smoothly change the current that is drawn from this DC voltage source, from 40mA to 20mV (we change the resistance from 225Ω to 450Ω). We create AC-current. But the DC voltage does not change (0Ω internal resistance). So the dynamic resistance of DC-voltage source is equal to rd = 0V/20mA = 0Ω And this is why we say that DC-Voltage is short for an AC-signals.

Here you have CS amplifier 






simulate this circuit &ndash; Schematic created using CircuitLab

And this is how his equivalent small-signal circuit looks like







simulate this circuit