Years ago, I swear I was able to make a high-side switch with a single P-channel Mosfet, even though my gate voltage was lower that the source voltage. I think I did it with a feedback from the drain. Am I crazy?
Was the resistor you have shown connected between the gate and drain perhaps connected between the gate and source instead? Then, grounding the gate will turn it on. Letting the gate be pulled high turns it off.
When drawing schematics involving MOSFETs, I always try to have the gate terminal on the left.
If it's a P-ch MOSFET, I prefer having the source terminal towards the top of the schematic, and the drain towards the bottom.
If it's an N-ch MOSFET, I prefer having the source terminal down, drain up.
This goes along with the standard practices of having inputs coming from the left, outputs towards the right, more positive voltages near the top, and more negative near the bottom of the schematic. It also makes it much easier to grasp what's going on in the schematic.
Here is a minimal parts example:
R1 keeps the gate at the same voltage of the source terminal when S1 is open. This keeps current from flowing from the drain terminal.
When S1 is closed, the gate is discharged via R2. R2 might seem to be unnecessary, but when S1 is closed, the inductance of the wiring and capacitance of the gate would form an LC circuit, and oscillate at a high frequency, which could cause the MOSFET to heat. R2 "snubs" this oscillation quickly.