Cheap proof that the Sorgenfrey line is normal?

Suppose $A,B$ are disjoint closed sets in the Sorgenfrey line. We have to find open sets $U,V$ such that $A\subseteq U,\ B\subseteq V,$ and $U\cap V=\emptyset.$

For each $a\in A$ choose $a'\gt a,$ $\ [a,a')\cap B=\emptyset$; then $U=\bigcup\limits_{a\in A}[a,a')$ is an open set containing $A.$

For each $b\in B$ choose $b'\gt b,$ $\ [b,b')\cap A=\emptyset$; then $V=\bigcup_\limits{b\in B}[b,b')$ is an open set containing $B.$

To show that $U\cap V=\emptyset,$ it suffices to show that $[a,a')\cap[b,b')=\emptyset$ for all $a\in A,\ b\in B.$ Suppose $a\in A,\ b\in B,$ and assume without loss of generality that $a\lt b.$ Since $[a,a')\cap B=\emptyset,$ it follows that $b\ge a'$ and so $[a,a')\cap[b,b')=\emptyset.$

That argument is as easy as any that I know, and its second part is an important result in its own right. An alternative is to prove that every linearly ordered topological space is hereditarily normal and then show that the Sorgenfrey line is a subspace of a linearly ordered space. The second step is easy. Let $X=\Bbb R\times\{0,1\}$ ordered lexicographically: $\langle x,i\rangle\preceq\langle y,j\rangle$ if and only if $x<y$, or $x=y$ and $i\le j$. Give $X$ the order topology generated by $\preceq$, and let $Y$ be the subspace $\Bbb R\times\{1\}$ with the topology that it inherits from $X$. If $\Bbb S$ denotes the Sorgenfrey line, the map

$$f:\Bbb S\to Y:x\mapsto\langle x,1\rangle$$

is easily seen to be a homeomorphism. $Y$ is normal as a subspace of the linearly ordered space $X$, so $\Bbb S$ is normal.