The circuit will work as you've drawn it, provided you get the diodes the right way round, but it won't be very efficient... about two thirds of the energy taken from the battery will be wasted in warming the resistors (220 ohms each is required to limit the current to about 25mA).
Little PP3 style 9v batteries don't have a huge capacity, so you could arrange things to be more efficient by having two parallel chains, each consisting of two diodes and a resistor in series. That way you'll take half as much current from the battery and have the same amount of light. Resistors in that case would each be 82 ohms.
Wire and switch requirements aren't a problem, because the currents and voltages are tiny.
![[Image: LEDs.jpg]](http://s10.postimage.org/jlrbm1xc9/LEDs.jpg)
Little PP3 style 9v batteries don't have a huge capacity, so you could arrange things to be more efficient by having two parallel chains, each consisting of two diodes and a resistor in series. That way you'll take half as much current from the battery and have the same amount of light. Resistors in that case would each be 82 ohms.
Wire and switch requirements aren't a problem, because the currents and voltages are tiny.