The following is my attempted proof that the sequence $\{ \frac{1}{n} \}_n$ converges in the Sorgenfrey line. Offer criticism, please!
Consider $\{\frac{1}{n}\}_n=(1,\frac{1}{2},\frac{1}{3},\cdots)$. If $\{\frac{1}{n}\}_n$ converges, then it must be that $\{\frac{1}{n}\}_n \rightarrow 0$. By definition this means that for each $W \in \tau$ containing $0$ there is an $N \in \mathbb{N}$ such that $x_n \in W$ for all $n \geq N$.
If it does not converge, then there is a $W \in \tau$ containing $0$ such that for each $N \in \mathbb{N}$ there is an $n \geq N$ with $x_n \notin W$.
Case 1. $W = [a,b)$ where $a < 0 < b$.
Since $b > 0$, by the Archimedean property there is an $n \in \mathbb{N}$ with $b>\frac{1}{n}$, so $0 < \frac{1}{n} < b$. This implies $b > \frac{1}{n} > \frac{1}{n+1} > \frac{1}{n+2} > \cdots > 0$. So $\frac{1}{m} \in W$ for all $m \geq n$.
Case 2. $W = [a,b)$ where $a = 0 < b$.
Since $b > 0$, by the Archimedean property there is an $n \in \mathbb{N}$ with $b>\frac{1}{n}$, so $0 < \frac{1}{n} < b$. This implies $b > \frac{1}{n} > \frac{1}{n+1} > \frac{1}{n+2} > \cdots > 0$. So $\frac{1}{m} \in W$ for all $m\geq n$.
It is a lot more instructive and easy to show convergence directly. There is no need for the claim "if the sequence converges, then it converges to $0$". Simply go ahead and show the limit is $0$. That is, given an open set $W$ containing $0$, by definition of the Sorgenfrey line there exists $a<b\in \mathbb R$ with $0\in [a,b)\subseteq W$. Thus $0<b$. Now take $n$ large enough so that $\frac{1}{n}<b$. Conclude your result.