proof of opamp's inputs being equal
It's quite clear.

The equation for the inverting input voltage for that circuit is 

$$ V_- = V_{IN}\left(\dfrac{R_F}{R_{IN}+R_F}\right) + V_{OUT}\left(\dfrac{R_{IN}}{R_{IN}+R_F}\right) $$

The equation for the output voltage for that circuit is

$$ V_{OUT} = -\dfrac{R_F}{R_{IN}}  V_{IN} $$

So, the first equation above becomes

$$ V_- = V_{IN}\left(\dfrac{R_F}{R_{IN}+R_F}\right) -\dfrac{R_F}{R_{IN}}  V_{IN} \left(\dfrac{R_{IN}}{R_{IN}+R_F}\right)  = 0$$

But really, there's a far simpler way to "see" the result that the voltage at the inputs of an ideal op-amp must be equal when negative feedback is present.

(1) The output voltage of an ideal op-amp is \$A(V_+ - V_-)\$ in the limit as \$A \rightarrow \infty \$.  Or as often said, the gain of an ideal op-amp is "infinite".

(2) Thus, for the output voltage to be finite, \$(V_+ - V_-) \$ must go to zero as \$A \rightarrow \infty \$, i.e., the input voltages must become exactly equal in the limit of infinite gain.

(1) and (2) establish that, for an ideal op-amp, the input voltages must be equal for the output voltage to be finite.

What is left to show is that negative feedback enforces this constraint while positive feed back opposes it which is not done in the linked answer.