Q:

Why does the second law of thermodynamics prevent 100% efficiency?

So far in my thermodynamics lecture course, my understanding of the laws of thermodynamics is that the first law is about the conservation of energy, the second law says entropy must always increase or stay the same which apparently results in the fact you can never achieve 100% efficiency of heat engines, unless at $T = 0\,\mathrm K$, and the last law says that you can't get to $T= 0\,\mathrm K$.
I have never explicitly seen why the fact that entropy must always increase or stay the same results in the prevention of achieving 100% efficiency. The only proof I have is showing the Carnot cycle is the most efficient and that is only 100% efficient if the cold reservoir is at absolute zero, which it can not be at. 
Is there any way to work from the statement: $\Delta S \geq 0$ (for any process in a closed system), to some result which says you can not achieve 100% efficiency?

A:

If you can convert all of the heat to work, you're reducing entropy by definition ($\Delta S = \frac{Q}{T}$ , If $Q<0$ then $\Delta S < 0$). 
If you allow yourself to let some heat flow into somewhere cold (heating something up instead of using all of the heat to work) you raise the entropy in the cold substance enough to let you not defy the second law, and the rest can go to useful work.

A:

The fact that a heat engine cannot be 100% efficient is a consequence of the Kelvin-Plank statement of the second law, which can be summarized as
Kelvin-Plank Statement of Second Law
"No heat engine can operate in a cycle while transferring heat with a single heat reservoir" (my emphasis on cycle)
COROLLARY to Kelvin-PLank: No heat engine can have a higher efficiency than a Carnot Cycle operating between the same reservoirs.
You are probably aware that in the Carnot reversible isothermal expansion the heat provided by the high temperature reservoir equals the work done in the expansion. That process converts heat to work at 100% efficiency. But a process is not a cycle. To operate in a cycle the system has to return to its original state. After the reversible isothermal expansion there is no way to return to the original state and do net work without rejecting some heat to a lower temperature  reservoir.
You could, for example, immediately follow the reversible isothermal expansion with a reversible isothermal compression and return to the original state. But the negative compression work would equal the positive expansion work for no net work done by the "cycle". 
In order to produce net work in the Carnot cycle it is necessary to follow the isothermal expansion with an reversible adiabatic expansion. That reduces the temperature of the system to that of the lower temperature reservoir. Then a reversible isothermal compression then followed by a reversible adiabatic compression returns the system and surroundings to their original states and net work is done. But that work equals the heat into the system minus heat rejected, for an efficiency of less than 100%.
All heat engine cycles must reject some heat for an efficiency of less than 100%. The Carnot cycle does this in the most efficient manner.

I understand this, I guess what I am having trouble seeing is how the
  statement Δ≥0 is equivalent to the kelvin plank statement. 

Actually we can show that the only way we can complete a cycle and have $\Delta S_{system}=0$ is to reject heat to a low temperature reservoir. By definition, a thermodynamic cycle is one where all the system properties (entropy, internal energy, pressure, temperature, etc.) are returned to their original state. 
So let us transfer heat from the surroundings to a system by means of a reversible isothermal expansion of an ideal gas. Since for an ideal gas $\Delta U$ only depends on temperature, from the first law $\Delta U=0$ and $W=Q$, and theoretically we have the complete conversion of heat to work at 100% efficiency for the process. However, during the isothermal expansion the entropy of the system has increased by $\Delta S_{sys}=+\frac{Q_H}{T_H}$ where $Q_H$ is the heat transfer from the high temperature reservoir $T_H$. 
In order to complete the cycle we must return the entropy of the system to its original state, meaning we must get rid of entropy of the amount $\frac{Q_H}{T_H}$. Now, here is the key point. The only way to transfer entropy is by heat transfer. That means to return the system to its original state we must transfer heat $Q_L$ to a low temperature reservoir. When we do this, the net work done in the cycle is
$$W_{net}=Q_{H}-Q_{L}$$ 
The efficiency $ζ$ of any cycle is the net work done divided by the gross heat added, or
$$ζ=\frac{W_{net}}{Q_H}=\frac{Q_{H}-Q_{L}}{Q_H}$$
Therefore, for any cycle,
$$ζ<1$$
The above applies to any cycle. It is whether or not a cycle is reversible that determines the maximum efficiency of the cycle. The maximum efficiency occurs when the cycle is reversible. For a reversible cycle both $\Delta S_{sys}$ and $\Delta S_{sur}$ are zero for $\Delta S_{tot}=0$. For an irreversible cycle there is additional entropy generated in the system which must be transferred to the surroundings as additional heat resulting in $\Delta S_{sur}>0$, and reducing the amount of work that can be done with the same heat input. 
Finally, as stated in the Corollary to Kelvin-Plank, "No heat engine can have a higher efficiency than a Carnot Cycle operating between the same reservoirs", which means the Carnot cycle is the most efficient of all reversible cycles. 

I was looking for a more explicit mathematical result which shows you can
  not conserve energy and follow Δ≥0 at the same time.

You can conserve energy and have $\Delta S_{tot}$ ≥0 at the same time. It happens when the cycle is not reversible. 
The First Law (energy conservation) and the Second law are completely independent laws. A cycle in which a process (or processes) is irreversible complies with the first law but generates entropy that must be transferred to the surroundings as additional (to the reversible cycle) heat. That in turn means less of the heat taken in does work in an irreversible cycle than the same cycle carried out reversibly. Since the entropy (and all other properties) of the system is unchanged for a complete cycle, the heat transferred to the surroundings results in $\Delta S>0$ for the surroundings.  
Hope this helps.

