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 Differnce 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 local minimum in $\left[{z \,.\,.\, z+1}\right)$

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