How to prove that every compact subspace of the Sorgenfrey line is countable?
Compact Subspaces of Sorgenfrey Line – How to Prove They Are Countable
compactnessgeneral-topologysorgenfrey-line
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(1) Let $X \subseteq \mathbb R$. Let $\mathcal S$ be the subspace topology that $X$ inherits as a subspace of the Sorgenfrey line, and let $\mathcal E$ be the topology that $X$ inherits from $E^1$, the reals with the Euclidean topology. You know that $\langle X, \mathcal E \rangle$ is separable, so there is a countable $D_0 \subseteq X$ that is $\mathcal E$-dense in $X$. This $D_0$ is almost $\mathcal S$-dense in $X$ as well.
To see this, let $V = (x,y] \cap X$ (where $x,y \in \mathbb R$ with $x<y$ ) be a non-empty basic open set in $\langle X, \mathcal S \rangle$. Then either $(x,y) \cap X \ne \emptyset$, or $V = \{y\}$. $(x,y) \cap X \in \mathcal E$ - it's open in the Euclidean topology - so if it's non-empty, it contains a point of $D_0$, and therefore so does $V$. We only have a problem in the second case, when $V = \{y\}$. Clearly such a point $y$ is an isolated point in $X$, so it must belong to any dense subset of $X$. Let $D = D_0 \cup \{y \in X:y \text{ is an isolated point in }X\}$. It should be clear that $D$ is dense in $\langle X, \mathcal S \rangle$, so the only question is whether it's countable. This will be the case provided that $E = \{y \in X:y \text{ is an isolated point in }X\}$ is countable.
For each $y \in E$ there must be a real number $x_y<y$ such that $(x_y,y] \cap X = \{y\}$. Consider the intervals $(x_y,y]$ in $\mathbb R$; clearly they must be pairwise disjoint (otherwise one would contain at least two points of $X$), so each must contain a different rational number. There are only countably many rationals, so there are at most countably many intervals $(x_y,y]$ with $y \in E$, $E$ is therefore at most countable, and $D$ is a countable $\mathcal S$-dense subset of $X$.
(2) Yes, the argument that you suggest works. You can also use Theorems VIII.6.2(3) and VIII.6.3 to see that $\mathbb R^\omega$ is second countable and therefore Lindelöf. I'll have to think a bit about your last question.
Edit: Sam's already answered the question and given a reason. If you're interested in the details, which are a bit messy, look at Theorem 3 in this paper by A.H. Stone. If you take his uncountable set $\Lambda$ to be $\mathbb R$, the space $T$ in that theorem is $(\mathbb Z^+)^\mathbb R$, and it's easy to check that it's a closed subspace of $\mathbb R^\mathbb R$. Since normality is inherited by closed subspaces, non-normality of $T$ implies non-normality of $\mathbb R^\mathbb R$.
First note the following characterisation of hereditary Lindelöfness:
Fact. A topological space $X$ is hereditarily Lindelöf iff given any family $\mathcal{U}$ of open subsets of $X$ there is a countable $\mathcal{U}_0 \subseteq \mathcal{U}$ such that $\bigcup \mathcal{U}_0 = \bigcup \mathcal{U}$.
proof. ($\Rightarrow$) If $X$ is hereditarily Lindelöf, then any family $\mathcal{U}$ of open subsets of $X$ is a cover of $A = \bigcup \mathcal{U}$ by open subsets of $X$, and so there is a countable $\mathcal{U}_0 \subseteq \mathcal{U}$ such that $A \subseteq \bigcup \mathcal{U}_0 \subseteq \bigcup \mathcal{U}$.
($\Leftarrow$) If $X$ is not hereditarily Lindelöf, then there is a subset $A \subseteq X$ and a cover $\mathcal{U}$ of $A$ by open subsets of $X$ such that $A \not\subseteq \bigcup \mathcal{U}_0$ for any countable $\mathcal{U}_0 \subseteq \mathcal{U}$. Clearly, $\bigcup \mathcal{U}_0 \neq \bigcup \mathcal{U}$ for all countable $\mathcal{U}_0 \subseteq \mathcal{U}$. $\dashv$
By this it suffices to show that for any family $\mathcal{U}$ of open sets in the Sorgenfrey line there is a countable $\mathcal{U}_0 \subseteq \mathcal{U}$ such that $\bigcup \mathcal{U}_0 = \bigcup \mathcal{U}$.
For each $U \in \mathcal{U}$ consider $\mathrm{Int}_{\mathbb R} ( U )$, where the interior is taken with respect to the usual metric topology on $\mathbb R$. As $\mathbb{R}$ is second-countable (and hence is itself hereditarily Lindelöf) there is a countable $\{ U_i : i \in \mathbb{N} \} \subseteq \mathcal{U}$ such that $$\bigcup_{i \in \mathbb{N}} \mathrm{Int}_{\mathbb R} ( U_i ) = \bigcup \{ \mathrm{Int}_{\mathbb R} ( U ) : U \in \mathcal{U} \}.$$
Note that if $x \in \bigcup \{ U : U \in \mathcal{U} \} \setminus \bigcup_{i \in \mathbb N} U_i$ then in particular $x \notin \mathrm{Int}_{\mathbb R} ( U )$ for all $U \in \mathcal{U}$, and so let us consider $A = \bigcup \{ U : U \in \mathcal{U} \} \setminus \bigcup \{ \mathrm{Int}_{\mathbb R} ( U ) : U \in \mathcal{U} \}$. I claim that $A$ is countable. For each $x \in A$ there is a $\epsilon_x > 0$ such that $[ x , x+ \epsilon_x ) \subseteq U$ for some $U \in \mathcal{U}$, and note that $( x , x+ \epsilon_x ) \subseteq \mathrm{Int}_{\mathbb R} ( U )$, so $( x , x+ \epsilon_x ) \cap A = \emptyset$. It follows that $\{ ( x , x+\epsilon_x ) : x \in A \}$ is a pairwise disjoint family of open sets in the metric topology on $\mathbb{R}$ and is therefore countable.
For each $x \in A$ pick $U_x \in \mathcal{U}$ containing $x$. Then $\mathcal{U}_0 = \{ U_i : i \in \mathbb N \} \cup \{ U_x : x \in A \}$ is a countable subfamily of $\mathcal{U}$ with the same union.
Best Answer
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.).