Every coanalytic set can be written as a union of $\aleph_1$-many Borel sets. (For example, see Theorem 4.3.17 of Srivastava's book "A course on Borel sets".)

  Take a coanalytic non-Borel set $B \subseteq \mathbb{R}^d$. Then $B = \bigcup_{\alpha \in \omega_1} B_{\alpha}$ for some Borel sets $B_{\alpha}$. For each $\alpha < \omega_1$, set $C_{\alpha} = \bigcup_{\delta < \alpha} B_{\delta}$. Then, each $C_{\alpha}$ is Borel, since it is a countable union of Borel sets. On the other hand, $\bigcup_{\alpha \in \omega_1} C_{\alpha}=\bigcup_{\alpha \in \omega_1} B_{\alpha}=B$ is non-Borel.

Every coanalytic set can be written as a union of $\aleph_1$-many Borel sets. (For example, see Theorem 4.3.17 of Srivastava.) Take a coanalytic non-Borel set $B \subseteq \mathbb{R}^d$. Then $B = \bigcup_{\alpha < \omega_1} B_{\alpha}$ for some Borel sets $B_{\alpha}$. For each $\alpha < \omega_1$, set $C_{\alpha} = \bigcup_{\delta < \alpha} B_{\delta}$. Clearly each $C_{\alpha}$ is Borel since it is a countable union of Borel sets. On the other hand, $\bigcup_{\alpha < \omega_1} C_{\alpha}=\bigcup_{\alpha < \omega_1} B_{\alpha}=B$ is non-Borel.