Hi,
The method is fine. Here you'd need to sink current ie pull the output to gnd instead of sourcing current ie pulling current from +v to load. Here you have the LED anode connected to the +v and the cathode connected to the PIC pin. When the PIC outputs 0, it sinks current, ie, pulls current from +v to gnd via LED, turning it on. Since you're using LED and 330R resistance, there's no question of PIC not being able to supply current as it can do so upto 15-20mA in the IO line.
Hope this helps.
Tahmid.