Quoted from Cserold:
Thanks guys. So, is this the complete wrong approach to dropping voltage from 12v to supply an LED needing 5v, 0.125A? How can I do that?
As others mentioned - the method was wrong. You need a complete voltage divider circuit. An accurate and consistent voltage divider is difficult to built for most applications as the load current tends to vary... which changes the voltage drop across the resistor.
Another downside to this method. You state that you want to go from 12V to 5V with a load current of 0.125A. If going with resistors configured as a voltage divider - the resistor needs to drop 7 volts across it. If you pick the right resistance values then you have a second problem. A resistor dropping 7 volts across it and conducting 0.125 amps means the resistor is dissipating 7 * 0.125 = 0.875 watts. The resistors you show appear to be 0.25W resistors - these would pretty much turn into little space heaters for a brief period of time.
The easiest and highly reliable method would be as BarakandI listed above - just use a simple 7805 regulator. A small heatsink should be used such as a Aavid 507302. For this simple of a circuit, you can easily just mount it on a terminal strip such as the one shown below. Or you could go with a cool running pre-built switcher such as suggested by Rock914 or setzkor.