How can I prove that the Sorgenfrey line is a Lindelöf space?

Here's a simple direct proof, which works just as well for the Sorgenfrey topology as for the usual topology of the line.

Let $\mathcal U$ be a collection of Sorgenfrey-open sets that covers $\mathbb R$. Let's say that a set $X\subseteq R$ is countably covered if $X$ is covered by countably many members of $\mathcal U$. We want to show that $\mathbb R$ is countably covered.

Consider any $a\in\mathbb R$, and let $C_a=\{x: x\ge a,\text{ and the interval }[a,x]\text{ is countably covered}\}$. It's easy to see that $\sup C_a=\infty$; assuming the contrary leads to a contradiction. Hence every finite interval $[a,b]$ is countably covered, and so is $\mathbb R=\bigcup_{n\in\mathbb N}[-n,n]$.

P.S. I have been asked to explain why assuming that $\sup C_a=b\in\mathbb R$ leads to a contradiction. Let $b_n=b-\frac{b-a}{2^n}$ for $n=1,2,3,\dots,$ so that $a\lt b_n\lt b$ and $b_n\to b.$ Thus for each $n$ there is a countable collection $\mathcal S_n\subseteq\mathcal U$ such that $[a,b_n]$ is covered by $\mathcal S_n,$ and the half-open interval $[a,b)$ is covered by the countable collection $\bigcup_{n\in\mathbb N}\mathcal S_n.$ Moreover, since $\mathcal U$ covers $\mathbb R,$ there is some $U\in\mathcal U$ such that $b\in U.$ Since $U$ is Sorgenfrey-open, there is some neighborhood $[b,b+\varepsilon)$ of $b$ (with $\varepsilon\gt0$) such that $[b,b+\varepsilon)\subseteq U.$ Then $[a,b+\varepsilon)$ is covered by $\{U\}\cup\bigcup_{n\in\mathbb N}\mathcal S_n,$ whence $b+\frac\varepsilon2\in C_a,$ contradicting our assumption that $b=\sup C_a.$

The proof given by bof is correct, but incomplete. Here is a completion.

When bof assumed ad absurdum that $\sup{C_a} = b \in \mathbb{R}$, he implicitely assumed that $b > a$, which a priori does not need to be true (as $b$ could be equal to $a$). For the sake of completeness, first remark that $C_a$ is non-empty, as the interval $[a, a]$ is trivially countably covered, so $a \in C_a$.

Now let $A$ be a Sorgenfrey-open set that covers $a$. Then, because $A$ is open, there must exist a neighbourhood $[a, a+\epsilon[$ in $A$ for some $\epsilon > 0$. Now observe that the interval $[a, a+\frac{\epsilon}{2}]$ is covered by $A$, so $a + \frac{\epsilon}{2}$ will certainly be a member of $C_a$. It follows that $b = \sup{C_a} \geq a + \frac{\epsilon}{2} > a$, which justifies bofs assumption that $b > a$.