If we introduce a finite temperature difference between source/drain and that of the working fluid in thermal contact, we would make the process irreversible since the direction of heat flow cannot reverse as it is only dependant on the temperature difference. This means that the direction of $Q_H/Q_L$ cannot change.
If we also ensure (for the forward cycle) that the working fluid traces the same path as a carnot cycle, the work done as well as the heat taken and lost matches that of the carnot engine. The process would merely be faster than a carnot cycle. Thus we would have an irreversible engine that has same efficiency of a reversible engine (which shouldn't be possible according to second law of thermodynamics), hence is also one that violates the following variation of Clausius' theorem:
$S_{irreversible} > \int \frac{\delta Q}T$
Where am I wrong in this?