that LED would be brighter, and cheaper
if you get that one - the resistor you need is worked out like this :-
The current supplied to the LED will be the supply voltage minus the LED forward voltage divided by the resistor value.
I = V / R were I is the current in amps, V is the voltage in volts, and R is the resistance in ohms.
We need to move the equation about as we are trying to work out the resistance R. V and I we already know from the LED specifications, and the supply voltage. If you are good a maths, you'll know that the equation can be moved around to R = V / I. It has been a long time since I did the maths on this, your maths teacher could explain how it gets moved around to this.
The forward voltage of that LED is 3.6V. the max forward current is 30mA, or 0.03A. For safety though, we should run it at 25mA or 0.025A, otherwise it could burn out if the voltage peaks a little higher. Now the resistor needs to drop the voltage from 5.8V to 3.6V at 0.025A.
this gives the equation R = Vbatt -Vled / I
this is the battery voltage minus the LED voltage, divided by the current
R = 5.8V - 3.6V / 0.025A
R = 2.2V / 0.025A
R = 88 ohms.
the nearest standard value for this 91ohms, but we could get away with going down to an 82ohm resistor - this will be slightly brighter, and will not burn out the LED at full charge.
I = Vbatt - Vled / R
I = 5.8V - 3.6V / 82
I = 2.2V / 82
I = 0.0268A, or 26.8mA - still under the 30mA limit
This chart will show you standard resistor values.
Now question is - can you tell me, from the resistor color chart before, what the color code will be, if we choose a resistor with 5% tolerance?
As for the battery holder - take the tin along with you, because I think the ones with the built in switch will be too big for the tin. That's why I recommended the other one, the dimensions will just squeeze into the tin. This one may be better -
Hopefully this isn't hurting your head by now!
Let me know how you get on.