I always use Jones' lemma. It's a handy tool to show non-normality of other spaces as well. You need some basic facts:
Suppose $X$ is normal. For every pair $A$, $B$ of closed disjoint non-empty subsets in $X$, there is a continuous function $f: X \rightarrow [0,1]$ such that $f(x) = 0$ for $x \in A$ and $f(x) = 1$ for $x \in B$. This is often called Urysohn's lemma.
If $f,g: X \rightarrow Y$ are continuous, and $Y$ is Hausdorff, and for some dense subset $D$ of $X$ we have $f(x) = g(x)$ for all $x \in D$, then $f(x) = g(x)$ for all $x \in X$. (Proof sketch: if not for some $x$, pull back disjoint open neighbourhoods of $f(x)$ and $g(x)$, both of these intersect $D$ and $f$ and $g$ cannot agree on those points.)
This implies:
2') The function $R$ that maps a continuous function $f$ from $X$ to $Y$ to a
continuous function $R(f)$ from $D$ to $Y$ by restricting $f$ to $D$, is 1-1.
Now,
Jones' Lemma: If $X$ is normal and $D$ is dense and infinite in $X$ and $C$ is closed and discrete (in the subspace topology) in $X$ then (as cardinal numbers) $2^{|C|} \le 2^{|D|}$.
Proof: for every non-trivial subset $A$ of $C$, $A$ and $C \setminus A$ are disjoint, closed in $X$ (both are closed in $C$, as $C$ is discrete, and closed subsets of a closed set are closed in the large set.), so by 1) there is a continuous function $f_A$ on $X$ that maps $A$ to $0$ and $C \setminus A$ to $1$.
Note that this defines a family of distinct continuous functions (if $A \neq B$ then we
can find a point in $A\setminus B$ or $B \setminus A$ that shows that $f_A \neq f_B$) from
$X$ to $[0,1]$. But from 2' we know that there is a 1-1 mapping from the set of all continuous functions from $X$ to $[0,1]$ to the set of all continuous functions from $D$ to $[0,1]$ and the latter set is bounded in size by $[0,1]^D = (2^{|N|})^{D} = 2^{|N||D|} = 2^{|D|}$, and the last step holds as $D$ is infinite.
As we have a family of size $2^{|C|}$ (all non-trivial, i.e. non-empty, non-$C$, subsets of $C$) we conclude that $2^{|C|} \le 2^{|D|}$, and this concludes the proof.
Applications:
a) The Sorgenfrey plane: using the antidiagonal $C = \{(x, -x): x \in \mathbb{R} \}$ and $D = \mathbb{Q} \times \mathbb{Q}$ as dense subset. As $2^{|C|} = 2^\mathfrak{c} > \mathfrak{c} = 2^{|D|}$, Jones' lemma says that $X$ cannot be normal.
b) The Niemytzki plane (or Moore plane) (see e.g. here) is not normal, with a similar computation, using $C$ the $x$-axis and $D$ the rational points in the upper halfplane.
A closed subspace of a Lindelöf space is Lindelöf too, so if the Sorgenfrey plane is, so is $L$. But $L$ in the subspace topology is discrete, as every singleton is open as $[a,b) \times [-a, d) \cap L = \{(a,-a)\}$.
A discrete space is Lindelöf iff is is countable (take the open cover by singletons). $L$ is not countable, contradiction.
Best Answer
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 $[a,a')\cap[b,b')=\emptyset.$