It is pretty straight forward:
You didn't build the circuit as given in the example. What you have built is a different circuit entirely.
You have the voltage divider on the output of the amplifier instead of the input.
The point of the original circuit was to set the voltage using the divider and then use the opamp to provide current so as to keep the output steady when loaded. The output voltage is then pretty much independent of the load (as long as the opamp can supply enough current.)
Opamp output current is usually measured milliamperes - you shouldn't expect to power a microprocessor from it.
Your circuit is using the the amplifier to provide the voltage and then using the divider to lower it. That means the voltage will depend on the load - you could just leave out the amplifier because it isn't doing anything useful.
Connect R1 and R2 in series as they are now, but from +5V to ground.
Connect the junction of R1 and R2 to pin 5 of the opamp.
You ought to get 3.3V out at pin 7.
This is more or less how it ought to be, except that the input to the voltage divider goes to +5V:
This is what you built:
The first has the voltage divider at the input to the opamp.
Yours has the voltage divider at the output. This is incorrect for reasons explained above.