Sorgenfrey line is hereditarily separable How to prove that the Sorgenfrey line is hereditarily separable?
 A: This has a proof that for any ordered space separable implies hereditarily separable.
And the Sorgenfrey line is a subspace of a separable ordered space (e.g. the double arrow).
A: Hint: Let $S$ be a subset of the Sorgenfrey line. From each interval of the form $[q,r)$ in the original line, where $q,r \in \mathbb{Q}$, pick one point from $S$ if possible. Then characterize the points in $S$ that are not limits of the points you just chose. 
A: Let $A \subset ℝ$ be arbitrary. If $A$ is countable, then $A$ itself is dense and the proof is done. Assume that $|A| > |ℕ|$.
We will construct sets $D_1, D_2$ and show that their sum $D$ is countable and dense in $A$ (i.e. $D \cap U \neq ∅$ for all nonempty open sets $U$ in induced topology $\tau_S|A$). This will prove, that the Sorgenfrey line is hereditarily separable.
Let $\mathcal{B} = \{ [p, q) \cap A: p,q \in \mathbb{Q}, p < q \} \setminus ∅$.
Let $D_1$ be a set that contains one element from each $B \in \mathcal{B}$ (say "hello" to the Axiom of choice). Note that $D_1$ is a infinite countable set.
Remark. In fact, in the context of Euclidean topology the set $D_1$ would be already the one we are looking for. However, in the Sorgenfrey line, it is not. Consider $A = [0, π]$. The singleton $\{π\} = [π, 4) \cap A$ is open in $\tau_S|A$, but it might be the case that we had a bad luck and $D_1 = [0, π] \cap \mathbb{Q}$, so $D_1 \cap \{π\} = ∅$. Which means $D_1$ is not dense in $A$. We need a set $D_2$ to deal with such cases.
Let $ D_2 = \{A \cap \sup B : B \in \mathcal{B} \} \setminus ∅$.
Let $D_1 \cup D_2 = D$. Since $D_2$ is countable, set $D$ is countable as well. Now we only need to show, that $D$ is dense in $A$. Take aribtrary nonempty basic set $[x, y) \cap A$. It suffices to show, that $([x,y) \cap A) \cap D \neq ∅$. There are two cases:
I. $[x,y) \cap A = \{ r \}$. Take $p, q \in \mathbb{Q}$ such that $p \leq x \leq r < q < y$. We have
\begin{align}
\sup ([p,q) \cap A) &= \sup (\overbrace{([x,q) \cap A)}^{\ni r} \cup (\overbrace{[q, y) \cap A}^{= ∅})) \\
&= \sup ([x,y) \cap A) \\
&= \sup \{ r \} \\
&= \{r\}.
\end{align}
By construction $r \in D_2 \subset D$.
II. $\{ r,z \} \in [x,y) \cap A$. Without loss of generality assume $r < z$. Take $p, q \in \mathbb{Q}$ such that $x < p < z < q < y$.
Since the set $[p,q) \cap A$ is not empty ($z$ is there!), by construction, there exists $d \in D_1 \subset D$ such that
$$
d \in [p, q) \cap A \subset [x, y) \cap A.
$$
q.e.d.
