For an isolated system, for both reversible and irreversible process
dQ = 0. So according to the equation of entropy isn't dS always zero
for an isolated system?
Although a differential change in entropy is defined for a reversible transfer of heat, or
entropy can be generated without heat transfer. Any irreversible process can generate (increase) entropy.
The classic example given is an ideal gas located in one side of a rigid insulated vessel with a vacuum in the other side separated by a rigid partition. Since the vessel is both rigid and insulated, the gas is an isolated system. There is no work or heat transfer between the gas and the surroundings.
An opening is created in the partition allowing the gas to expand into the evacuated half of the vessel. $W=0$, $Q=0$, $\Delta T=0$ (for an ideal gas) and therefore $\Delta U=0$. Although no heat transfer has occurred, the process is obviously irreversible (you would not expect the gas to be able to spontaneously return to its original location) and entropy increases.
You can calculate the entropy increase by assuming any convenient reversible process that can bring the system back to its original state (original entropy) and apply the above definition for entropy. We can do this because entropy is a state function that does not depend on the path.
The obvious choice is to remove the insulation and insert a movable piston. Then conduct a reversible isothermal compression until the gas is returned to its original volume leaving a vacuum in the other half. All properties are then returned to their original state. The change in entropy for the isothermal compression is then, where $Q$ is the heat transferred to the surroundings by the isothermal compression,
Since the system is returned to its original state, the overall change in entropy is zero, meaning the original change in entropy due to the irreversible expansion had to be
Hope this helps.