How to prove that compact subspaces of the Sorgenfrey line are countable?

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.).

Hint: any uncountable set of real numbers contains a strictly increasing infinite sequence.

Hint 2 (added later): show that if a subspace $X$ of the Sorgenfrey line contains a strictly increasing infinite sequence, then $X$ has an open cover with no finite subcover.