Here is a simple way.
You know that in order for the Carnot Cycle to be reversible, both the isothermal expansion and compression must be carried out quasi-statically, that is, very slowly. That means the temperature difference between the system (say an ideal gas) and the surroundings during the isothermal processes must approach zero in the limit. Lets take the isothermal expansion as an example.
Let the temperature of the surroundings be $T_{surr}$ and the temperature of the gas be $T_{sys}$. Let heat $Q$ be isothermally transferred from the surroundings to the system. The changes in entropy thus become:
$$\Delta S_{surr}=-\frac{Q}{T_{surr}}$$
It is negative because heat is transferred out of the surroundings. Now the change in entropy of the system becomes:
$$\Delta S_{sys}=+\frac{Q}{T_{sys}}$$
The total entropy change becomes:
$$\Delta S_{tot}=+\frac{Q}{T_{sys}}-\frac{Q}{T_{surr}}$$
Now note that for any $T_{surr}>T_{sys}$, $\Delta S_{tot}>0$. Only when the temperature of the system equals the temperature of the surroundings in the limit, will the total change in entropy approach zero. This is the case for a reversible heat transfer.
But we know that for any real heat transfer process to occur, there needs to be a temperature difference, and so the total entropy change will always be greater than zero. So all real processes are irreversible. The Carnot cycle establishes an upper limit to the efficiency of a cycle operating between two thermal reservoirs.
Hope this helps.