Chemistry - t-butylbenzene in highly acidic medium
Solution 1:
All 6 positions on the benzene ring are being protonated (the frequency of protonation at the different sites is related to the stability of the carbocation generated) in acidic media, so all of the ions you suggested are being formed. Protonation at 5 of these positions can only regenerate starting material when one of the two protons at the initial site of protonation is removed (an identity reaction, we could make this pathway visible by running the reaction in deuterio-acid + $\ce{D2O}$). Protonation of the benzene ring at the carbon bearing the substituent provides a choice. Either the proton can again be removed or the stable t-butyl cation can be ejected. All of these things are happening in the reaction, but only removal of the t-butyl group is visible.
Solution 2:
I think the problem you are having stems from locally optimizing the energy, but disregarding the global energy well that you get in the end.
The mechanism your book gives is valid because it explains how this reaction could take place along a minimal-energy pathway. Since the reaction is observed, it somehow has to proceed from tert-butylbenzene to the products benzene and tert-butanol.
I know that this is a slight tongue-in-cheeky kind of argument, but in the end it is a valid one.
If you protonate elsewhere (and we haven't discussed the protonation kinetics involved here1 but they might be important) you will not end up cleaving the right $\ce{C-C}$ bond (or any bond at all). The tert-butyl cation is quite stable, and having stable reaction products means that the reaction is more happy to proceed. Since the protonation is likely to be a fast equilibrium reaction, it will happen at many sites on the aryl group.
1 Frankly I have no clue about those, so maybe someone can help me out here...