Mechanism of acid-catalyzed Robinson annulation?
As I mentioned in my Comment the only serious issue was the formation of an allylic anion in acid medium. You seem to know how to form enols from ketones and how to form ketones from enols. I will ignore those steps in the diagram. Protonation of the carbonyl in 1 can cause ring fragmentation to form transient enol 2 that rapidly tautomerizes to diketone 3. Cyclization via enol 4 leads to β-hydroxyketone 5. Although loss of water from 5 as you have described is viable, I prefer formation of the enol 6 because loss of water is facilitated over a tertiary cation given the allylic nature of the species 7. Loss of a proton from carbocation 7 leads to 3-methylcyclohex-2-en-1-one (8).
Addendum: The process is not a Robinson annulation, which involves an initial Michael addition followed by an aldol condensation. The latter reaction includes the transformation 3 $\rightarrow$ 8. The transformation 1 $\rightarrow$ 3 may described as a retro-aldol reaction.
You may want to see the formation of mesitylene (1,3,5-trimethylbenzene) from acetone under acid conditions.