Sorgenfrey Line is First-Countable

Theorem
Let $\mathbb R$ be the set of real numbers.

Let $\mathcal B = \{[a \,. \, . \, b) | a,b \in \mathbb R\}$.

Let $\tau$ be the topology generated by $\mathcal B$, that is, the Sorgenfrey line.

Then $\tau$ is first-countable.

Proof
Let $\mathcal B_x = \{ \left[x \,. \, . \, x + \frac{1}{n}\right) | n \in \mathbb{N}_{>0}\}$.

By definition of first-countability, it suffices to show that:


 * (1): $\mathcal B_x$ is countable
 * (2): $\mathcal B_x$ is a local basis at $x$

(1) follows from the fact that $\mathcal B_x$ is a bijection from the set of natural numbers.

(2):

By definition of local basis, it suffices to show that $\forall U \in \tau: x \in U: \exists B \in \mathcal B_x: B\subseteq U$.

Pick any $U$ in $\tau$.

By definition of $\tau$, there exists $[x \,. \, . x+\varepsilon)\subseteq U$ for some $\varepsilon \in \mathbb{R}_{>0}$.

By Archimedean Principle there exists $n\in \mathbb N$ such that $n > \dfrac{1}{\varepsilon}$ (i.e. $\dfrac{1}{n} < \varepsilon$).

So $x \in \left[x \,. \, . \, x+\dfrac{1}{n}\right) \subseteq \left[x \, . \, . \, x+\varepsilon\right) \subseteq U$.