Virtual Ground Paradox?

It's #2. For a "perfect" theoretical opamp, the open-loop gain is infinite, and this makes the difference at the inputs zero. When introducing opamp circuits, or when working out how things are supposed to work, people normally think about the "perfect" opamp.

When thinking about the performance of a circuit, we usually have to start thinking about the imperfections of a real opamp. For a real opamp, the open-loop gain is not infinite, and there is some difference between the inputs. To take the example of an LM324, the open loop gain is about 115dB. That's a little less than a million volts/volt, so if there is a 1V DC output, then the inputs are different by about 1uV. Most of the time you can ignore that.

It gets more complicated for AC. At higher frequencies, the gain drops. For the LM324, it goes to 0dB, i.e. 1V/V at about 1MHz. At that point, the inputs certainly will have a large difference. Practically speaking, the amplifier just doesn't work any more. For frequencies in-between, the gain of the amplifier (inc. feedback) will vary. The term "Gain Bandwidth Product" is used to describe what gain you can have at what frequency for a given opamp.

This is just one of many imperfections a real opamp has. Another very relevant one is input offset voltage. This is the difference in inputs which results in a zero output, and it's not always exactly 0. This might be more important than the limited gain in many cases. Other imperfections you might want to consider are saturation/clipping, input current, PSRR, CMRR, nonzero output impedance and many more.

The problem is that you mix-up two different models of the op-amp.

A real, but somewhat idealized op-amp, is a differential amplifier whose output depends on the inputs as follows (neglecting saturation):

$$ V_{out} = A_{Vol} \cdot (V^+ - V^-) $$

Using this simplified model (simplified because it neglects saturation, offset voltage, bias currents, bandwidth and other real-world effects) and the fact that \$A_{Vol}\$ (open loop gain) is huge, you can prove that, when the op-amp is connected in a negative feedback circuit, then the virtual short circuit holds, but only when you approximate \$A_{Vol}\$ as infinite.

With this drastic approximation you can have a zero differential input AND still a finite output, since the open loop gain is assumed infinite.

In reality the open loop gain isn't infinite and your finite output is due to a very small differential input (in the μV range, usually). Multiply that small differential input by the actual open loop gain and you have your finite output.

Using the virtual short circuit, is much simpler, though. Once you realize that an op-amp circuit has negative feedback, you can use the virtual short circuit idealization (\$V^+ = V^-\$) to analyze how the circuit works, without bothering with the actual value of the differential input, which becomes irrelevant (unless you need the finer details), as long as you avoid output saturation.

Let's just do the WHOLE shebang, start to finish, instead of doing this piecemeal. Let's start with the definition for the op amp.

$$ V_{out}= A_{OL}(V_+ - V_-)$$

As has been pointed out, \$A_{OL}\$ is a very big number, but let's leave it in place for the time being.

Just converting this into the notation in the original figure, $$ V_{B}= A_{OL}(0 - V_A)$$ $$ V_B=-V_AA_{OL}$$

Now, we can start applying Kirchoff's Current Law.

$$\frac{V_{in}-V_A}{R_{in}}= \frac{V_A-V_B}{R_f}$$

$$\frac{R_f}{R_{in}}(V_{in}-V_A)=V_A-V_B$$

$$ V_B=V_A - \frac{R_f}{R_{in}}(V_{in}-V_A)$$

$$ V_B = V_A \left( 1 + \frac{R_f}{R_{in}} \right)- \frac{R_f}{R_{in}}V_{in}$$

Now, we can substitute in for \$V_A\$, based on the definition of the op amp $$ V_B = -\frac{V_B}{A_{OL}} \left( 1 + \frac{R_f}{R_{in}} \right)- \frac{R_f}{R_{in}}V_{in}$$

Lastly, now we can apply \$A_{OL}\to\infty \$ , which makes the first term go to zero.

$$\lim_{A_{OL}\to\infty} V_B = - \frac{R_f}{R_{in}}V_{in}$$

This is your standard inverting amplifier equation. Also, note that \$V_A=-\frac{V_B}{A_{OL}}=0\$, leaving us with a "virtual ground" at the inverting input. Thus, there is no paradox. The virtual ground concept is entirely consistent with an infinite open loop gain op amp in a negative feedback arrangement. For giggles, try the same exercise in positive feedback, and watch it explode.

Carrying these things through without throwing out terms due to assumptions also shows you where errors are likely to crop up. For example, you can see from the equation before taking the limit that if you're asking for obscene gain, and \$R_f\$ is many orders of magnitude larger than \$R_{in}\$ that things may not work out so well.