Is it consistent that there exists a partition $P$ of the real number line $\mathbb{R}$ such that $|P|>\aleph_{0}$ but where $\bigcup R$ is Borel whenever $R\subseteq P$?
If $2^{\aleph_{0}}<2^{\aleph_{1}}$, then the answer to this question is $\textbf{no}$ since there would be at least $2^{\aleph_{1}}$ subsets of each uncountable partition $P$ but there are only $2^{\aleph_{0}}$ Borel sets. We therefore need to use a model such that $2^{\aleph_{0}}=2^{\aleph_{1}}$ to construct our counterexample.
Under $MA+\neg CH$ does there exist a partition $P$ of the real number line $\mathbb{R}$ such that $|P|>\aleph_{0}$ but where $\bigcup R$ is Borel whenever $R\subseteq P$?
Perhaps such a partition $P$ is impossible to construct by some fact obvious to descriptive set theorists that I probably have overlooked.
I call a partition $p$ of a Boolean algebra $B$ such that $\bigvee R$ exists whenever $R\subseteq p$ a subcomplete partition and subcomplete partitions came up all the time in my work on Boolean partition algebras, so I would be interested in which partitions of a given Boolean algebra are subcomplete and which ones are not.