Can't charge a cell phone with a power bank I built
You have three problems:


The 7805 requires at least 7&nbsp;volts (7805 datasheet) on its input to function correctly. A fully charged 6&nbsp;volt battery is some where around 6.4&nbsp;volts, so there is not enough for the regulator to work reliably.
Diode D1 eats 0.7&nbsp;volts, so your 7805 is getting less than 6&nbsp;volts - it is definitely not enough to work right.
R1 (regardless of value) shouldn't be there. It will reduce the current flow, and you will have a voltage drop across it.


So, your battery voltage is too low, you are losing voltage across D1 and R1, and the current limiting is not necessary and is contributing to your problems.

To use the 6-volt battery, you could use just R1 and D1 in series. D1 will drop the battery voltage down to about 5.7&nbsp;volts, then the series resistor will reduce the current a bit so that the phone's charge can live with the over voltage. R1 would be around 1&nbsp;ohm (0.7) if you need 1&nbsp;ampere for the charger, around 1.5 to 2&nbsp;ohms for 500 milliamperes. You would need resistors capable of handling 1&nbsp;watt (better, 2&nbsp;watts) of power. If you take eight resistors rated for 1/4&nbsp;watt and a value of 10&nbsp;ohms each and put them in parallel, then you would have 2&nbsp;watts of power capability and in the neighborhood of 1 to 2&nbsp;ohms (1.25 by calculation, the real value depends on the precision of the resistors.)

Alternatively, you could look into finding a low dropout (LDO) regulator that can work with less than 1.5&nbsp;volts difference between input and output. The 2940 (datasheet) is a simple, three pin LDO regulator that can provide 5&nbsp;volts from 5.5&nbsp;volts.  Note: The 2940 looks like the 7805, but it has a different pinout than the 7805, so you can't just slap it in and be happy.