Why acting on a closed system infinitely slowly does not increase its entropy? I don't understand why a finite-time process necessarily increases entropy while an infinite-time process might not.
 A: As pointed out by Mr. Knighton, the entropy (S) of a system is a state function just like temperature (T), pressure (P), internal energy (U), enthalpy (H), etc., which means that in going from equilibrium state 1 to equilibrium state 2 the entropy at state 2 of the system (and thus the change in entropy of the system) does not depend on the process (path function) that gets you there, be it fast or slow, reversible or irreversible.  However, the total change in entropy (system + surroundings) will depend on the process. This is where Veronika’s question becomes relevant.
To give a concrete example, consider a system H (a hot body) and its surroundings C (a cold body). Further consider both H and C to be thermal reservoirs, that is, they are so massive that a heat transfer between them doesn’t change their temperatures. The temperature of H is TH and the temperature of C is TC. We bring the bodies together and desire to transfer heat Q from H to C. Since the temperature of either does not change, the heat transfers occurs isothermally.  Let’s look at the entropy changes:
For Body A (System):  ΔSA = -Q/TH (a drop in entropy)
For Body B (Surroundings):  ΔSB = +Q/TC (a rise in entropy)
The total entropy change:  ΔSTot = ΔSA + ΔSB  
Then for any TH > TC you will see that:      Q/TC - Q/TH > 0 
In order for the total entropy change to approach zero, the temperature difference must approach zero.  This results in the heat transfer rate approaching zero and the time it takes to transfer Q infinitely long. In order for the total entropy change to actually equal zero, the temperatures would have to be the same- but if that were the case we would have no heat transfer at all! 
Sorry if this answer was too long.
