Thevenin Impedance

If you think about it that VCVS is not dependent. It depends on some parameters that are not part of the circuit you are analyzing, so you can turn it off (better than saying "short it" in my opinion.

You can also see this from another point of view: if you include the \$V^+V^-\$ part of the circuit and connect a generator to these terminals, then you should turn it off, then \$A(V^+-V^-)=0\$. Or you can just say that if you only have dependent sources in a circuit, then the "all nodes are at 0V" solution works, and some backend math tells us that in a nice network the solution is uniqe if it exists, so also if you don't include the generator on the op amp input terminals you can say well the output must be zero.

Understood this the solution is trivial and it's precisley what your book reports.

Your analysis is correct.

We first replace the current source with an open circuit and the voltage source with a short circuit.

Now we note that the bottom 2k resistor is in series with the capacitor but this series combination is shorted by the voltage source so they can be removed from the circuit in this simplification.

Now we have a 2k resistor in series with a short circuit so the impedance is in this case purely resistive and has a value of 2k as you thought.

Note its important to keep thinking of this as an impedance, while in this circuit the impedance is purely resistive, most ac circuits with inductors or capacitors in them will have an impedance that varies with frequency and has a phase shift associated with it.