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.
how does the first resistor's resistance influences the second
resistor's resistance.
It doesn't. However, the resistance of the first resistor influences the voltage across the second resistor.
Clearly, the resistors in the diagram are series connected thus the current through each resistor is identical.
$$I_1 = I_2 = I$$
By Ohm's law, we have
$$V_1 = I_1 \cdot R_1 = I \cdot R_1$$
$$V_2 = I_2 \cdot R_2 = I \cdot R_2$$
By KVL, we have
$$6V = V_1 + V_2 = I \cdot R_1 + I \cdot R_2 = I \cdot (R_1 + R_2)$$
Thus
$$I = \frac{6V}{R_1 + R_2}$$
In other words, the current \$I\$ depends on the sum of the resistances (series connected resistances add).
The voltage across the second resistor can now be written as
$$V_2 = I \cdot R_2 = \frac{6V}{R_1 + R_2} \cdot R_2 = 6V \frac{R_2}{R_1 + R_2} $$
and so, as first stated, the resistance of the first resistor influences the voltage across the second resistor.
Similarly
$$V_1 = I \cdot R_1 = \frac{6V}{R_1 + R_2} \cdot R_1 = 6V \frac{R_1}{R_1 + R_2}$$
Best Answer
Of course there is, but it does not look pretty. Make the divisors equal, and add the terms. for three resistors you get
\$ \dfrac{R2 \cdot R3}{R1 \cdot R2 \cdot R3} + \dfrac{R1 \cdot R3}{R1 \cdot R2 \cdot R3} + \dfrac{R1 \cdot R2}{R1 \cdot R2 \cdot R3}\$
\$ = \dfrac{(R2 \cdot R3) + (R1 \cdot R3) + (R1 \cdot R2)}{R1 \cdot R2 \cdot R3} \$
now do the \$ \dfrac{1}{n} \$ and you get:
\$ \dfrac{R1 \cdot R2 \cdot R3}{(R2 \cdot R3) + (R1 \cdot R3) + (R1 \cdot R2)} \$
The top line is easy, it is the product (multiplication) of all resistors. The bottom line is the sum of the products of all leave-one-out combinations. For two that reduces to the pretty formula:
\$ \dfrac{(R1 \cdot R2)}{(R1+R2)}\$