Electronic – Why are these resistors between different nodes assumed to be parallel

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 circuit you analyse with your formula is essentially this one, with the switch open: two independent braches, each with two resistors in series.

^{simulate this circuit – Schematic created using CircuitLab}

The current in the two braches is independennt, each current is V / ( R + R ).

Note that the potential (voltage) at both sides of the switch is the same, so we can close the switch, without effect on the circuit. Now we have your circuit, except that R1 is represented by TWO parallel resistors, each R, so the equivalent is R/2.

To summarize, you analyzed your circuit as if it were the one I show, which is different from your circuit in the value of R1.

The comment from The Photon gives another view, which amounts to the same thing.