Maybe my misunderstanding of topology and compact sets is seriously flawed, but as far as I can tell I can't see why the "Sorgenfrey Line" is not $\sigma$-compact.
A topological space is $\sigma$-compact if it is the countable union of compact subsets. The Sorgenfrey Line is simply the real line endowed with the topology of half-open intervals. As stated on Wikipedia, the sets $(-\infty, a)$ and $[b, \infty)$ are both open, implying $[a, b)$ is clopen, and hence compact for finite $a, b$ (closed and bounded, unless Heine-Borel doesn't hold in this space for some reason).
Then we can write $\mathbb{R} = \bigcup_{n\in\mathbb{Z}}[n-\epsilon,n+1)$ for some $\epsilon>0$, so the Sorgenfrey line is clearly $\sigma$-compact.
What's the flaw in the logic? Is it the assumption that Heine-Borel holds in this space, and so the subsets with finite open subcovers aren't exactly the closed, bounded sets?
Heine-Borel is specific to the Euclidean topology, and need not hold in a general space (where "bounded" makes no sense, necessarily), and it does not hold for the Sorgenfrey line: $[0,1]$ can be covered by the open sets $[0,1-\frac{1}{n})$, $n \in \mathbb{N}$ and $[1,2)$ and this cover has no finite subcover. So $[0,1]$ is not compact.
In fact, I showed here that a compact subset of the Sorgenfrey line is at most countable. So a $\sigma$-compact subset is also at most countable.
As a bonus an indirect proof: suppose $S$ were $\sigma$-compact. Then $S \times S$ would be $\sigma$-compact (hence Lindelöf) as well, and $T_3$, and a $T_3$ Lindelöf space is normal while $S \times S$ is not normal.
Or, $D':= \{(x,-x) : x \in S\} \subset S \times S$ is closed, uncountable and discrete and so $S \times S$ cannot be $\sigma$-compact, as otherwise $D'$ would be too, which it is not: the only $\sigma$-compact discrete spaces are the countable ones.