Using Open drain to get +5V
An open drain pin can only sink current, so you are right about the pullup resistor, but a transistor is not necessary. Also Vss is the ground pin which should be at 0V:



Vdd is the positive supply pin, and Vee is the negative supply pin (as mentioned in the 4502 datasheet Vee can be connected to Vss/circuit ground if you don't have a negative rail)  

Internally, the open drain pin looks like the part in the dotted rectangle:



Simulation - notice how the MOSFET is sinking the current through the resistor to pull the pin to 0V intially, then when the MOSFET shuts off at 100ms, the resistor pulls the line back up to 5V (in this case - could be +3.3V, or another supply voltage within specs of the ICs involved)



So all you need is the pullup resistor to the 4052 positive supply rail. This can be anywhere between, say, 2.2k&Omega; and 100k&Omega;. Generally for something that only switches slowly and speed of switching is not important, something like 10k&Omega;-22k&Omega; would do fine.
The higher the resistance, the more susceptible to the line being affected by noise, and the slower the line switches, so if the PCB is in a noisy environment (or there is a lot of high current/high speed switching on the PCB itself) you may want to choose a lower value (the downside is increased power consumption when the line is pulled low)