How does a diode-connected MOS device 'clamp' the voltage
Can't Vgs be greater than Vth for a diode connected deivce?, so then the Vdd to Vx difference will no longer be a constant Vth.

That's not the point here. The point is, for a given voltage difference between the gates of M1 and M2 there will be defined currents through M1 and M2 (sharing Iss). This current through M1 sets the \$V_{gs}\$ of M3 and thereby the voltage drop from \$V_{DD}\$ to point X.
So for a given input the voltage the voltage on point X is always \$V_{DD} - V_{gs}\$.