I understand that for a discrete symmetry breakdown where the wave functional for your field has two vacua, say $|+\rangle$ and $|-\rangle$, the reason why the true groundstate can't be a superposition of these two is because they are separated by a potential barrier and, assuming the field extends over an infinite volume, the total energy barrier is infinite. So tunneling is impossible.
But let's say you have a continuous symmetry and a Mexican hat potential. Then you have infinitely many vacua, denoted by $|\theta\rangle$ where $0\le\theta<2\pi$. If I pick two of these states, say $|0\rangle$ and $|\varepsilon\rangle$ where $\varepsilon$ is infinitesimal, the potential barrier between them is zero: the two states have the same energy and are next to each other, so tunneling should be permitted. And then once you have tunneled from $|0\rangle$ to $|\varepsilon\rangle$, you can tunnel to $|2\varepsilon\rangle$ then $|3\varepsilon\rangle$ and so on all the way back to $|0\rangle$. You will then have a superposition of all such states, and the superposition would respect the symmetry.
Except that doesn't happen. Where's the flaw in the reasoning?