Ordered union of Borel sets Let $\mathfrak{A}$ be an uncountable collection of Borel sets in $\mathbb{R}^d$ such that for any $A,B\in\mathfrak{A}$, either $A\subset B$ or $A\supset B$. Then is it necessarily true that the union $\bigcup_{A\in\mathfrak{A}}A$ is Borel measurable?
 A: 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.
A: 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.
