So why is this a valid proof for all resistors in parallel
First, you have an error in your question - the equivalent resistance is
$$R_P = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$$
Now, the voltage across the two parallel resistors is what it is regardless of how the voltage comes to be.
However we choose to label that voltage is immaterial, thus, we can arbitrarily label the voltage across the parallel resistors as, e.g., \$v_P\$.
Now, and again, it does not matter how this voltage comes to be, the voltage variable \$v_P\$ is the voltage measured across the parallel resistors when "red" lead is placed on the "\$+\$" labelled terminal and the "black" lead is on the "\$-\$" labelled terminal.
Thus, by Ohm's law, the current through each resistor is
$$i_{R_1} = \frac{v_P}{R_1} $$
$$i_{R_2} = \frac{v_P}{R_2} $$
So, the total current is, by KCL,
$$i_P = i_{R_1} + i_{R_2}$$
and the equivalent resistance is defined as
$$R_P = \frac{v_P}{i_P}$$
thus,
$$R_P = \frac{v_P}{i_{R_1} + i_{R_2}} = \frac{v_P}{\frac{v_P}{R_1} + \frac{v_P}{R_2}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$$
Again, if we replace the two parallel resistors with a resistor of resistance \$R_P\$, the current through the equivalent resistance will be identical to the sum of the currents through the two parallel resistors.
The 20mA rating of your LEDs is probably the recommended maximum current. Most LEDs will produce light at much lower currents - I find 5 mA is sufficient for indicator light use.
Try the LEDs with different resistors to see what current is actually required to produce sufficient light for your application (more current = brighter light).
Also, check the maximum current the Pi GPIO pins can handle - I have a feeling it is less than 20 mA, but can't check at the moment.
Best Answer
It could be one of three reasons:
Increasing power handling capabilities of a single resistor. If the footprint only allows for a certain size package (say, for height reasons), multiple resistors will spread the wattage out.
A pick & place machine is assembling the circuit, and its cheaper to parallel 4 resistors in one spot than have another reel with the right value.
It's homework and they want you to do math.
As for resistance decreasing in parallel, in this instance you can treat resistors as cashiers at the grocery store. If you have one cashier who can scan 50 items a minute, and you add a second who can scan 50 items per minute, the total number of items scanned is 100 per minute, instead of 25. Since items per minute approximates electron flow, adding more resistors side by side will allow more electrons to flow.