Prove that a Change of Volume will not Change the Value of the Equilibrium Constant
In your workyou have changed the "equilibrium constant" which is a fixed while keeping the number of moles constant of all species in both instances. Rather its the opposite

Consider the reaction;$$\ce{AB &lt;=&gt; A + B }$$ with equilibrium constant $K_c$ at temperature $T$ and volume $V_0$

Initially let $n(AB) = x_0$ , $n(A) = y_0$ , $n(B) = z_0$ where $n()$ refers to the number of moles.

then $$K_c =  {\frac{{\frac {x_0}{V_0}}{\frac {y_0}{V_0}}}{{\frac {z_0}{V_0}}}}$$

Now the volume is reduuced from $\ce{V_0\bond{-&gt;}V}$
Hence all the concentrations increase.
Then the reaction quotient at the moment the volume is changed;$$Q_c =  {\frac{{\frac {x_0}{V}}{\frac {y_0}{V}}}{{\frac {z_0}{V}}}}$$

Now $Q_c &gt; K_c$ hence equilibrium shifts to the right reducing the number of moles of $A$ and $B$ increasing the number of moles of $AB$.Due to this effect $Q_c$ will gradually reduce and at acertain point will becaome equal tp $K_c$ with different number of moles.

$$K_c =  {\frac{{\frac {x_0}{V_0}}{\frac {y_0}{V_0}}}{{\frac {z_0}{V_0}}}} = {\frac{{\frac {x}{V}}{\frac {y}{V}}}{{\frac {z}{V}}}} $$

The only parameter remaining constant(under constant temperature) is the Equilibrium constant is a certain parameter(i.e. Volume here) is changed, other parameters will vary keeping Equilibrium constant a constant.