Is the voltage divider in this circuit backwards for what I'm attempting to achieve?
The problem was that I needed to lower the voltage on the base pin of the BJT so the user suggested I add the voltage divider (R2, R3)


That's not really what's going on here, but I guess that's what you're asking about.

To use a voltage divider to lower the applied voltage you'd need a resistor in series with the switch, so that when the switch closed it would be between the supply node and the divider node.


  The original target was to apply 0.5 - 0.7 V on the base pin.


Your circuit as drawn should already achieve this, because the transistor itself will limit the voltage on the base pin to ~0.5 - 0.7 V, if you limit the amount of current supplied (which you are doing due to R2)

In the circuit you have, R3 is really just providing a path to discharge the base node when the switch is open. In a MOSFET circuit this would be important if you needed a quick shut-off after opening the switch. In a BJT circuit it's not as important since current can be sunk through the base-emitter junction. But it will still somewhat speed up the shut off. It will also waste about 2.5 mW when the switch is closed.

Edit

Now that you've linked the original question, I can point out that the answer you got there never said this was a voltage divider.

What he said was,


  R2 will limit the base current when thet switch is closed. R3 pulls the base low when the switch is open, to ensure the transistor is not conducting.


Which is exactly the same as what I'm telling you now.

You don't need a voltage divider to reduce the voltage at the base of the BJT. If you limit the current (using R2), the properties of the BJT will take care of the voltage for you.