Weight of Sorgenfrey Line is Continuum

Theorem
Let $T = \left({\R, \tau}\right)$ be the Sorgenfrey Line.

Then $w \left({T}\right) = \mathfrak c$

where
 * $w \left({T}\right)$ denotes the weight of $T$
 * $\mathfrak c$ denotes continuum, the cardinality of real numbers.

Proof
By definition of Sorgenfrey Line, the set:
 * $\mathcal B = \left\{{\left[{x \,.\,.\, y}\right): x, y \in \R \land x < y}\right\}$

is a basis of $T$.

By definition of weight:
 * $w \left({T}\right) \leq \left\vert{\mathcal B}\right\vert$

where $\left\vert{\mathcal B}\right\vert$ denotes the cardinality of $\mathcal B$.

By Cardinality of Basis of Sorgenfrey Line not greater than Continuum:
 * $\left\vert{\mathcal B}\right\vert \leq \mathfrak c$

Thus
 * $w \left({T}\right) \leq \mathfrak c$

It remains to show that:
 * $\mathfrak c \leq w \left({T}\right)$

Aiming for a contradiction, suppose that
 * $\mathfrak c \not\leq w \left({T}\right)$

Then:
 * $w \left({T}\right) < \mathfrak c$

By definition of weight, there exists a basis $\mathcal B_0$ of $T$:
 * $w \left({T}\right) = \left\vert{\mathcal B_0}\right\vert$

Then by Set of Subset of Reals with Cardinality less than Continuum has not Interval in Union Closure:
 * $\exists x, y \in \R: x < y \land \left[{x \,.\,.\, y}\right) \notin \left\{{\bigcup A: A \subseteq \mathcal B_0}\right\} = \tau$

By definition of $\mathcal B$:
 * $\left[{x \,.\,.\, y}\right) \in \mathcal B \subseteq \tau$

Thus this contradicts by definition of subset with:
 * $\left[{x \,.\,.\, y}\right) \notin \tau$