Can you turn on a NMOS with a lower voltage than at the drain?
What actually matters for making a MOSFET conduct or not is not \$V_{gs}\$ !

What matters is if there is a channel formed under the gate. This depends on the gate-body voltage.

On discrete MOSFETs the source and the body are shorted



That makes the drain different is it is not shorted to the body. Also note that there is a pn junction (diode) between body and drain. This diode is always present both on drain-body and on source-body. This also means that we have to keep the drain voltage higher than the body = source voltage otherwise the drain-body diode would start to conduct. If that happens current can flow whatever the gate voltage is !

As long as the source = body is at the lowest voltage (compared to the drain) then that voltage determines if there is a conductive channel.

soon as the FET starts to conduct, the voltage present at the source will become the drain voltage, in this case, 3V.

That's not true when used in a circuit where the source is grounded, when the MOSFET is used as a switch the Drain voltage should be pulled down to become equal to the source voltage. Not the other way round.

Also in the case you describe, if the source voltage does become higher there will be a point when \$V_{gs} = V_{th}\$ and then what happens ? Yes the MOSFET stops conduction. So how can the source voltage increase even further then ?

What you describe is what happens in a source follower circuit:
 

Which is not a good circuit for switching something on/off. Instead it is used to buffer a voltage while introducing a voltage shift (of roughly \$V_{th}\$).