Set of Subset of Reals with Cardinality less than Continuum has not Interval in Union Closure

Theorem
Let $\BB$ be a set of subsets of $\R$, the set of all real numbers.

Let:
 * $\card \BB < \mathfrak c$

where
 * $\card \BB$ denotes the cardinality of $\BB$
 * $\mathfrak c = \card \R$ denotes continuum.

Let $\FF = \set {\bigcup \GG: \GG \subseteq \BB}$.

Then:
 * $\exists x, y \in \R: x < y \land \hointr x y \notin \FF$

Proof
Define:
 * $ Z = \leftset {x \in \R: \exists U \in \FF: x}$ is local minimum in $\rightset U$

By Set of Subsets of Reals with Cardinality less than Continuum Cardinality of Local Minimums of Union Closure less than Continuum:
 * $\card Z < \mathfrak c$

Then by Cardinalities form Inequality implies Difference is Nonempty:
 * $\R \setminus Z \ne \O$

Hence by definition of empty set:
 * $\exists z: z \in \R \setminus Z$

By definition of difference:
 * $z \in \R \land z \notin Z$

Thus $z < z + 1$.

We will show that $z$ is local minimum in $\hointr z {z + 1}$.

Thus:
 * $z \in \hointr z {z + 1}$

Hence:
 * $z - 1 < z$

Thus:
 * $\openint {z - 1} z \cap \hointr z {z + 1} = \O$

Then by definition $z$ is a local minimum in $\hointr z {z + 1}$.

Because $z \notin Z$:
 * $\hointr z {z + 1} \notin \FF$