Driving MOSFET gate using PIC

What is wrong with your circuit is irrelevant from the point of view of getting a working unit.

There is no need for a MOSFET driver in this case. Use the IRLML2502 MOSFET and drive its gate directly from the digital output of the microcontroller. Check the IRLML2502 datasheet and you will see that it has a maximum guaranteed on resistance of only 45 mΩ with 4.5 V on the gate. You also don't need a resistor in series with the gate:

Yes, it really is that simple.

The reason the FET turns off so slowly is because you only have 2 kΩ pulling it high. Take a look at the datasheet for that FET. It should show you the actual and effective gate capacitance when switching. The relatively weak 2 kΩ pullup is working against that capacitance.

Here is a trick I sometimes use in this situation:

The double emitter follower is basically a impedance buffer. The current to charge and discharge the gate capacitance is handled by Q1 and Q2, so the signal at GATE can be much higher impedance and still switch the FET quickly. Note that this double emitter follower will loose 700 mV or so on each end. That is fine for a FET that switches over a 10 V gate range. It will still be "off" with 700 mV on the gate just as well as with 0. At the other end, its usually easy enough to drive GATE the extra 700 mV further than what you want to make sure shows up on the gate of Q3.