I think the intended question was about a thermally isolated system, so no heat could be exchanged; work exchange is allowed.
What's wrong with the reasoning here?
In short, the statistical entropy that is defined for any equilibrium state and can increase in thermally isolated system is not the same thing as the information entropy - a functional of the probability density (or phase space volume where the probability density is the same everywhere) that has to remain constant due to the Liouville theorem. These two entropies are related, but they are not the same concept nor do they always have the same value.
1)
The second law says that entropy can only increase
More accurately, it says that when state is changed from equilibrium state A to equilibrium state B, thermodynamic entropy cannot decrease. Equilibrium state can be specified by stating values of sufficient number of macroscopic variables $X_1,X_2,...$. (For example, equilibrium state of an ideal gas is specified by giving values of volume $V$ and internal energy $U$.) Then thermodynamic entropy function $S(\mathbf X)$ of these state variables can be introduced.
With this, the non-decrease of entropy can now be stated mathematically in this way:
$$
S(\mathbf X_B) \geq S(\mathbf X_A).
$$
2)
entropy is proportional to phase space volume.
More accurately, statistical entropy of an equilibrium state specified by macroscopic variables $X_1,X_2,...$ is proportional to logarithm of volume of certain phase space region. It is believed statistical entropy gives accurate replacement for thermodynamic entropy function (which is the subject of the 2nd law).
The phase space region used above is defined as follows. Any point of phase space represents possible state of the Hamiltonian system (so-called microstate). Not all microstates are compatible with the macroscopic variables $X_1,...$; those that are form the region. Let us denote volume of this region by $\Omega(\mathbf X)$. The statistical entropy of macrostate $\mathbf X$ is defined by
$$
S^{(stat)}(\mathbf X) = k_B \log \Omega(\mathbf X).
$$
Thus statistical entropy is a function of volume $\Omega$, which in turn is a function of the macroscopic variables $\mathbf X$. If the state changes, $\Omega$ may change; and if it does, so does $S^{(stat)}$.
3)
Liouville's theorem says that phase space volume is constant
That is true, but only within the right context. The volume that is sometimes used to formulate the Liouville theorem is a different thing from the volumes $\Omega(\mathbf X_A), \Omega(\mathbf X_B)$ used to define entropy.
The difference is that in the definition of statistical entropy $S^{(stat)}$, the volume $\Omega$ is a function of macroscopic variables $\mathbf X$; it is not a function of time. Consequently, statistical entropy is not a function of time. It does not evolve from lower to higher value continuously in time. It is not necessarily defined for all times between the time $t_1$ where the system is in state A and time $t_2$ where the system is in state B. It is only defined for equilibrium states; in our case, we consider only states A and B.
On the other hand, a popular way to state the Liouville theorem is:
Volume of moving points selected in past remains constant in time.
More accurately, the Liouville theorem implies that if at some time $t_1$ we select points forming a phase space region $R_1$ of definite volume $\Delta\omega_{R_1}$, then at some later time $t_2$, those same points may be elsewhere and form a region $R_2$ of entirely different shape, but its volume $\Delta\omega_{R_2}$ equals $\Delta\omega_{R_1}$. It follows that we can introduce a function of time $\Delta\omega(t)$ that gives volume of the region considered at any time as it moves. It is a constant function:
$$
\Delta\omega(t_1) = \Delta\omega(t) = \Delta\omega(t_2) ~~~\forall t\in\langle t_1;t_2\rangle.
$$
You may ask: couldn't we apply this to volume $\Omega(\mathbf X_A)$, so the observed volume $\Delta \omega$ would actually be equal to the volume of the whole phase space $\Omega(\mathbf X_A)$? Wouldn't it then follow that $\Omega$ cannot change and thus we must have $\Omega(\mathbf X_A) = \Omega(\mathbf X_B)$?
We could, as a special application of the Liouville theorem, do just that. As time goes by, the observed points will be moving and the region they define may change its shape.
$$
\Delta \omega(t_1) = \Omega(\mathbf X_A) = \Delta \omega(t)~~~\forall t\in\langle t_1;t_2\rangle.
$$
So indeed the phase space volume of observed points does not change in time. But isn't this a problem? What if we calculate the statistical entropy as
$$
S^{(stat)}(\mathbf X_B) = k_B \log \Delta \omega(t_2) ?
$$
This has to have the same value as
$$
S^{(stat)}(\mathbf X_A) = k_B \log \Delta \omega(t_1).
$$
But if so, this prevents having higher entropy in the final state B.
The actual problem is that the formula for statistical entropy is not, as some believe
$$
S^{(stat)}(t) = k_B\log \Delta\omega(t)
$$
where $\Delta\omega(t)$ is volume of phase space occupied by observed set of points obtained as a result of Hamiltonian evolution in time. That is a different notion, a special case of information entropy that is a functional of probability distribution, and thus may be a function of time. In this situation, where there is no heat transfer so the Liouville theorem applies, the information entropy remains constant in time, even if statistical entropy does not.
The correct formula for calculating statistical entropy is based solely on the macroscopic thermodynamic variables, not time. We know that by assumption, the system at time $t_2$ is in macroscopic state $\mathbf X_B$. The phase space volume corresponding to this state is $\Omega(\mathbf X_B)$. If it is greater than $\Omega(\mathbf X_A)$, the statistical entropy has increased. It is this phase volume based on macroscopic variables, not the volume of the observed points in the phase space, that is important for statistical entropy. The volume of phase space compatible with state B may be much higher than volume of phase space compatible with Hamiltonian evolution from A at time $t_1$ to B at time $t_2$.