On the complex plane $\mathbb C$ consider the half-open square $$\square=\{z\in\mathbb C:0\le\Re(z)<1,\;0\le\Im(z)<1\}.$$

Observe that for every $z\in \mathbb C$ and $r\in\{0,1,2,3\}$ the set $(z+i^r\cdot\square)$ is the shifted and rotated square $\square$ with a vertex at $z$.

Problem. Is it true that for any function $r:\mathbb C\to\{0,1,2,3\}$ there a subset $Z\subset\mathbb C$ such that the union of the squares $$\bigcup_{z\in Z}(z+i^{r(z)}\cdot\square)$$is not Borel in $\mathbb C$?

Added in Edit. As @YCor observed in his comment, the answer to this problem is affirmative under $\neg CH$.