basic p-type mosfet circuit, breaks mosfet?
First, the use of a mosfet is unnecessary to drive an enable pin of virtually any IC. One can simply use a circuit like this:



simulate this circuit &ndash; Schematic created using CircuitLab
Assuming the enable pin does not source or sink excessive current, when the switch is open, the enable pin will be at 0V and when the switch is closed, the enable pin will be at 1.66V (close to the 1.65V you intended). If the enable pin does source or sink significant current, the voltages will be be 0 and 1.66V. R1 and R2 may be decreased to compensate, (with the side effect that more power will be used when the switch is closed).
Second, the MOSFET you have chosen (like most mosfets) has a body diode connected between the source and drain pins. As configured, the drain (and hence Vout) will never rise beyond 0.6-0.7V above the source/ground.