Finding Thevenin Impedance of a circuit
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.