Construction
Consider some non-Borel set $Y \subset [0,1]$ (e.g. Vitali set).
Enumerate $Y$ using ordinals as $Y=\{x_\alpha\}_{\alpha < \beta}$. Let $m$ denote the smallest ordinal such that $X=\{x_\alpha\}_{\alpha < m}$ is non-Borel. Note that $m\ge \omega_1$, since otherwise $X$ would be at most countable.
Then the family $\mathfrak A = \{A_\gamma\}_{\gamma<m}$, where $A_\gamma = \{x_\alpha\}_{\alpha<\gamma}$, has the properties desired in the OP, but $\bigcup_{A \in \mathfrak A} A$ is non-Borel.
For more details about ordinals see e.g. Set theory by T. Jech (2006).
Discussion of cardinality
Let us prove that in fact $|X| = \aleph_1$.
Indeed, by construction $\aleph_1 \le |X| \le 2^{\aleph_0}$. Any uncountable Borel set has cardinality $2^{\aleph_0}$ (e.g. by Theorem 13.6 in Classical Descriptive Set Theory by A.S. Kechris or Corollary 2C.3 in Descriptive Set Theory by Y.M. Moschovakis.). Therefore, if CH fails then $|X|\le \aleph_1$, since otherwise there would exist a Borel subset with cardinality $\aleph_1$ (by minimality of $m$). On the other hand if CH holds then immediately $|X| = \aleph_1$.
Most of this answer comes from the very useful comments in this thread.