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.
Suppose $A,B$ are disjoint closed sets in the Sorgenfrey line. We have to find open sets $U,V$ such that $A\subseteq U,\ B\subseteq V,$ and $U\cap V=\emptyset.$
For each $a\in A$ choose $a'\gt a,$ $\ [a,a')\cap B=\emptyset$; then $U=\bigcup\limits_{a\in A}[a,a')$ is an open set containing $A.$
For each $b\in B$ choose $b'\gt b,$ $\ [b,b')\cap A=\emptyset$; then $V=\bigcup_\limits{b\in B}[b,b')$ is an open set containing $B.$
To show that $U\cap V=\emptyset,$ it suffices to show that $[a,a')\cap[b,b')=\emptyset$ for all $a\in A,\ b\in B.$ Suppose $a\in A,\ b\in B,$ and assume without loss of generality that $a\lt b.$ Since $[a,a')\cap B=\emptyset,$ it follows that $b\ge a'$ and so $[a,a')\cap[b,b')=\emptyset.$
Best Answer
$\mathcal{D}$ is discrete because for any $x$, the set $U_x:=[x,x+1) \times [-x,-x+1))$ is an open neighbourhood of $(x,-x)$ in $\mathcal{L}_S \times \mathcal{L}_S$ and it intersects $\mathcal{D}$ in $\{(x,-x)\}$, so all its points are isolated (this intersection is open in the subspace topology by definition).
That $\mathcal{D}$ is closed is clear; it's already closed in the coarser Euclidean topology on the plane.
The final contradiction comes from the fact that Jones' lemma implies $2^{2^{\aleph_0}} \le 2^{\aleph_0}$ which flat-out contradicts Cantor's theorem that for any cardinal $k$, $k < 2^k$.