For each $\alpha\in (0,1)\setminus \mathbb Q$ and $t\in \mathbb R$, consider the sequence $(a_n)_{n\ge 0}$ with $a_n=\lfloor \alpha n+t\rfloor - \lfloor \alpha (n-1)+t\rfloor$. This is called the Sturmian sequence with angle $\alpha$, an elements of $\{0,1\}^{\mathbb Z^+}$. Let $S_\alpha$ be the orbit closure under the shift of $(a_n)$. This is a closed invariant set under the shift (by definition). The frequency of 1's in blocks of elements of $S_\alpha$ converge uniformly to $\alpha$. Hence the $S_\alpha$ are disjoint.

Now you can turn these into invariant sets for the squaring map: $Z_\alpha=\{e^{2\pi i t}\colon t\in S_\alpha\}$ is an invariant set for the squaring map. These remain disjoint.

For each $\alpha\in (0,1)\setminus \mathbb Q$ and $t\in \mathbb R$, consider the sequence $(a_n)_{n\ge 0}$ with $a_n=\lfloor \alpha n+t\rfloor - \lfloor \alpha (n-1)+t\rfloor$. This is called the Sturmian sequence with angle $\alpha$, an elements of $\{0,1\}^{\mathbb Z^+}$. Let $S_\alpha$ be the orbit closure under the shift of $(a_n)$. This is a closed invariant set under the shift (by definition). The frequency of 1's in blocks of elements of $S_\alpha$ converge uniformly to $\alpha$. Hence the $S_\alpha$ are disjoint.

Now you can turn these into invariant sets for the squaring map: $Z_\alpha=\{e^{2\pi i t}\colon t\in S_\alpha\}$ is an invariant set for the squaring map. These remain disjoint.

PS: Not sure if minimal was added later, or if I missed it, but the $Z_\alpha$ are minimal also.