Infinite closed partition of the real numbers with a certain property Is there a partition of the real numbers into infinitely many closed subsets so that no infinite union of these subsets (other than the whole set of real numbers) is closed?
 A: Perhaps it's worthwhile to state a weaker result I was able to come up with.  There is a partition of $\mathbb R$ into sets of cardinality $2$ such that no uncountable union of these sets (other than $\mathbb R$) is closed.
The cardinality of the set $\cal F$ of closed proper subsets of $\mathbb R$ is that of the continuum, so
we can well-order it by the first ordinal of cardinality $\cal c$.  Thus we have a well-ordering $\prec$ of $\cal F$ in which for each $A \in \cal F$, the set of $B \in \cal F$ such that $B \prec A$ has cardinality strictly less than $\cal c$.  Now using transfinite recursion, for each $A \in \cal F$, take $P_A$ as follows.  If $A \backslash \bigcup_{B \prec A} P_B$ is nonempty, choose one of its members to be in $P_A$.  Also choose one or two members
of $A^c \backslash \bigcup_{B \prec A} P_B$ to be in $P_A$, so that $P_A$ has two members.
Note that the nonempty open set $A^c$ has cardinality $\cal c$, which is larger than the cardinality of $\bigcup_{B \prec A} P_B$, so this is always possible.  Any uncountable closed subset of $\mathbb R$ has cardinality $\cal c$, so if $A$ is uncountable $P_A$ will contain a member of $A$. 
By construction the $P_A$ are all disjoint.  Their union is all of $\mathbb R$, since any $x$ will be in  $P_{\{x\}}$ if it is not in some $P_B$ for $B \prec \{x\}$.  And  no uncountable closed set $C$ can be a union of the sets $P_A$: such a union would have to include $P_C$, because one member of $P_C$ is in $C$, but $P_C$ also contains a member of $C^c$.
