The effect of pressure on equilibrium constant
First step: Don't use Kc... use Kp!

You are focusing on concentration, but as we're working with gases it becomes easier to work with partial pressures.

$K_\text{p}= \frac{\ce{p(NO2)}^2}{\ce{p(O2)}\ce{p(NO)}^2}$

Now, if we remember that partial pressure is pressure $\times$ mole fraction ($p_{a}=P \times \chi_{a}$), and substitute these in we get (and cancelling out the pressures):

$K_\text{p}= \frac{\chi_{(NO_{2})}^2}{\chi_{(O_{2})}\chi_{(NO)}^{2}P}$

Now, if we consider increasing the volume. By doing so (and assuming that temperature is kept constant), then we have decreased the pressure (ideal gas law: PV=nRT).

Now we can see that although the number of moles changes (and hence the mole fractions), our equilibrium constant will remain unchanged thanks to the effect of pressure.

A good guide is here.