$\mathfrak{F}$ family of open subsets of $\Bbb{R}$ s.t. the disjoint union of $\mathfrak{F} = \Bbb{R}$.Prove that $card(\mathfrak{F}) \leq \aleph_0$ I read this problem on a calculus book:

Let $\mathfrak{F}$ be a family of nonempty open subsets of $\Bbb{R}$ such that $\bigcup \mathfrak{F} = \Bbb {R}$ and that $\forall A, B \in \mathfrak{F} \, : A \ne B, A \bigcap B = \emptyset$ (i.e. they are pairwise disjoint). Prove that $Card(\mathfrak{F}) \leq \aleph_0$ (i.e. $\mathfrak{F}$ is either a finite or a denumerable set).

I found the solution on the book I was reading, but now I have another question:
Prove that $\mathbb R$ is not a nontrivial disjoint union of open subsets.
This thread states that if $\Bbb{R} = A \bigsqcup B$, then at least one of A or B is not an open set. So the question is: how could the book prove the statement if no such family esists?
I sincerely don't know how to tackle the problem. Could you please give me some hints?
Thanks a lot.
 A: I think the book is confusing two facts: 
$\mathbb{R}$ is not a disjoint union of two non-empty open sets. This is a restatement of the fact that $\mathbb{R}$ is connected, which follows from the order-density and order-completeness of the reals.
Any family of pairwise disjoint non-empty open sets in $\mathbb{R}$ is at most countable. Such a family need not have the whole set as its union, so e.g. $\{(n - \frac{1}{2} ,n+\frac12): n \in \mathbb{Z} \}$ is such a family.
This fact follows from the idea that any such interval contains a (necessarily distinct) rational (which is a countable set).
A: One can pick a distinct rational in each set, which (since they're disjoint) means cardinality of family at most that of the positive integers.
BTW I don't think this can be done except in case the family has only one member, all reals. 
Edit: By a "nontrivial" disjoint union is meant there are more than one set in the union. So the book is just saying at that point that the only qualifying collection of open sets is the single set $(-\infty,\infty).$
