One can expand the comments by François G. Dorais and James Hanson using ordinals.

First, assuming weak CH, note that there exists a non-Borel set with cardinality $\aleph_1$.
Indeed, take some set $X \subset [0,1]$ such that $|X| = \aleph_1$. The cardinality of the power set $\mathscr P(X)$ is $2^{\aleph_1}$, while the cardinality of the Borel sigma-algebra $\mathfrak B(\mathbb R)$ is $2^{\mathfrak \aleph_0}$, see e.g. [here](https://math.stackexchange.com/questions/70880/cardinality-of-borel-sigma-algebra). Hence there exists $A \in \mathscr P(X) \setminus \mathfrak B(\mathbb R)$. 

Next one can enumerate $A$ using ordinals as $A = \{x_\alpha\}_{\alpha \in \omega_1}$. For any $\alpha \in \omega_1$ let $A_\alpha := \{x_\beta : \beta \in \alpha\}$. Then $A_\alpha$ is countable and hence Borel. Moreover, for all $\alpha,\beta \in \omega_1$ either $A_\alpha \subset A_\beta$ or $A_\beta \subset A_\alpha$.

For more details about ordinals see e.g. *Set theory* by T. Jech (2006).