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

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

Let:
 * $\left\vert{\mathcal B}\right\vert < \mathfrak c$

where
 * $\left\vert{\mathcal B}\right\vert$ denotes the cardinality of $\mathcal B$
 * $\mathfrak c = \left\vert{\R}\right\vert$ denotes continuum.

Then:
 * $\exists x, y \in \R: x < y \land \left[{x \,.\,.\, y}\right) \notin \left\{{\bigcup \mathcal G: \mathcal G \subseteq \mathcal B}\right\}$

Proof
Define:
 * $\mathcal F = \left\{{\bigcup \mathcal G: \mathcal G \subseteq \mathcal B}\right\}$

Define:
 * $ Z = \left\{{x \in \R: \exists U \in \mathcal F: x}\right.$ is local minimum in $\left.U\right\}$

By Set of Subsets of Reals with Cardinality less than Continuum Cardinality of Local Minimums of Union Closure less than Continuum:
 * $\left\vert{Z}\right\vert < \mathfrak c$

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

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 $\left[{z \,.\,.\, z+1}\right)$.

Thus:
 * $z \in \left[{z \,.\,.\, z+1}\right)$

Hence:
 * $z-1 < z$

Thus:
 * $\left({z-1 \,.\,.\, z}\right) \cap \left[{z \,.\,.\, z+1}\right) = \varnothing$

Then by definition $z$ is a local minimum in $\left[{z \,.\,.\, z+1}\right)$.

Because $z \notin Z$:
 * $\left[{z \,.\,.\, z+1}\right) \notin \mathcal F$