(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 $\mathbb R^\mathbb R$.
Best Answer
First note the following characterisation of hereditary Lindelöfness:
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.