Determining the more acidic proton in cyclopenta-1,3-diene
Figure c does not look like a real compound. If it represents the conjugate base of cyclopentadiene, it is lacking a negative sign, and all five protons around the ring are equivalent, so the question of whether the red or blue proton is more acidic makes no sense.

If Figure c represents cyclopentadiene, the acid, it does not clearly show that the red position really has 2 hydrogens. However, the double bond structure shown is consistent with 2 hydrogens in the red position, and the question becomes very simple: in cyclopentadiene, is a methylene proton or a vinylic proton more acidic? Removing a vinylic proton puts a negative charge in the localized sigma system - no aromaticity develops. However, removing one of the methylene protons causes that carbon to go from sp3 to sp2p, and the negative charge goes into the pi system where it can be delocalized over the 5-membered ring. And aromaticity is a characteristic feature of cyclopentadienide ion.