on your picture, you have the incoming power "POWER IN 9V AC" but on your diagram you label the power as "+12v" (ac or dc not specified)
feeding 9v ac into the regulator is going to make it rather hot, and should have blown out that big capacitor you have on the input
in the diagram, you show the LED being in series with the regulator, and a 330 ohm resistor connecting the positive rail to ground. On the breadboard, it appears the led and resistor are connected correctly, in parallel with the regulator.
On the PIC, Pin 4, RA5 / MCLR should be connected to the positive rail with a 10k resistor.
On the left side of the breadboard, there appears to be four wires leaving the circuit - where do these go? If I am to assume two of them are power coming in, and two of them are for the switch, I recommend moving the 0v (ground) connection from the incoming power directly to the ground terminal of the regulator, instead of having it travel around the entire breadboard before reaching the regulator. I also recommend installing a 0.1 to 0.33 uF capacitor at the input and output of the regulator.