Let $C$ be a compact subset of the Sorgenfrey line (so $X = \mathbb{R}$ with a base of open
sets of the form $[a,b)$, for $a < b$). The usual (order) topology on $\mathbb{R}$ is coarser (as all open intervals $(a,b)$ can be written as unions of Sorgenfrey-open sets $[a+\frac{1}{n}, b)$ for large enough $n$, so are Sorgenfrey-open as well) so $C$ is compact in the usual topology as well. This means in particular that $C$ is closed and bounded in the usual topology on $\mathbb{R}$.
Suppose that $x_0 < x_1 < x_2 < \ldots $ is a strictly increasing sequence in $C$, and let $c = \sup \{x_n: n =0,1,\ldots \}$, which exists and lies in $C$ by the above remarks. Also let $m = \min(C)$, which also exists by the same.
Then the sets $[c,\rightarrow)$ and $[m, x_0)$ (if non-empty), $[x_n, x_{n+1})$, for $n \ge 0$ form a disjoint countable cover of $C$, so we cannot omit a single member of it (we need $[x_n, x_{n+1})$ to cover $x_n$, e.g.), so there is no finite subcover of it that still covers $C$.
This contradicts that $C$ is compact.
We conclude that $C$ has no infinite strictly increasing sequences. Or otherwise put: $C$ in the reverse order (from the standard one) is well-ordered.
And so we have shown that every compact subset of $C$ corresponds to a well-ordered subset of $\mathbb{R}$ (by reversing the order, and note that the reals are order isomorphic to its reverse order). And all well-ordered subsets of $\mathbb{R}$ are (at most) countable (this follows from several arguments, including one using second countability, e.g.).
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 $[a,a')\cap[b,b')=\emptyset.$
Best Answer
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.