Creating required voltage across a device using basic electrical principles
The answer by Oliver is indeed correct. However, I will just add why you are seeing the results you are seeing when trying to power something through your divider. 

You are correct in saying that for a power supply of 10V DC, dividing it down via 2 equal resistors will give you 5V. we see that here: 







simulate this circuit &ndash; Schematic created using CircuitLab

This is because the current flowing through this would be 10/20k = 500uA. the voltage drop across a 10k resistor is: 500uA * 10k = 5V.

So, what would happen if you fitted a load on this? Let us see:







simulate this circuit

You have now added extra resistance on to the bottom resistor. Assume that the load resistance is 10k. This would make the bottom resistor 5k (resistors in parallel). inputting this new figure into our previous equations will give is a middle point voltage of 3.3V.

So you can see that the voltage division rule does not break, you have just adjusted the values. 

Adding a voltage regulator (for 5V the classic 7805) will be able to ensure you have a stable output voltage no matter what the load. This is usually done with an internal compensation circuit such as this:







simulate this circuit