$\mathbb{R}_l \times \mathbb{R}_l$ is not Lindelöf. Here is my proof that the Sorgenfrey plane is not Lindelöf. In particular I just took a guess by claiming that the line $L$ is homeomorphic to $\mathbb{R}_l$, to claim uncountability. Is this correct? Also is everything else correct with the proof?
Consider the subspace $L=\{(x,-x)|x \in \mathbb{R}_l\}$ of $\mathbb{R}_l \times \mathbb{R}_l$. Then $L$ is closed since $\mathbb{R}_l \times \mathbb{R}_l-L=\{(x,y)|y>-x\}\cup \{(x,y)|y<-x\}$ is the union of two open half planes hence open. Then $\{\mathbb{R}_l \times \mathbb{R}_l-L\}\cup\{[-x,a) \times [x,b)|x,a,b \in \mathbb{R}, x<a,x<b\}$ is an open cover for $\mathbb{R}_l \times \mathbb{R}_l$, and $L \cap \{[-x,a) \times [x,b)|x,a,b \in \mathbb{R}, x<a,x<b\}=\{(-x,x)\}$. So $L$ and $\{[-x,a) \times [x,b)|x,a,b \in \mathbb{R}, x<a,x<b\}$ intersect in exactly one point. Since $L$ is uncountable being homeomorphic to $\mathbb{R}_l$, no countable subcollection of $\{\mathbb{R}_l \times \mathbb{R}_l-L\} \cup \{[-x,a) \times [x,b)|x,a,b \in \mathbb{R}, x<a,x<b\}$ covers $\mathbb{R}_l \times \mathbb{R}_l$.
 A: $L$ is definitely not homeomorphic to $\Bbb R_\ell$: $L$ is actually a closed, discrete subset of $\Bbb R_\ell\times\Bbb R_\ell$. In fact, what is essentially part of your argument actually shows that $L$ is discrete: for each $\langle x,-x\rangle\in L$, $[x,x+1)\times[-x,-x+1)$ is an open set in $\Bbb R_\ell\times\Bbb R_\ell$ whose intersection with $L$ is $\{\langle x,-x\rangle\}$, so $\{\langle x,-x\rangle\}$. $\Bbb R_\ell$, however, definitely does not have the discrete topology, so it is not homeomorphic to $L$. What is true is that the map $\Bbb R\to L:x\mapsto\langle x,-x\rangle$ is a bijection, so $|L|=|\Bbb R|$, and $L$ is indeed uncountable.
Assuming that you have already proved that a closed subspace of a Lindelöf space is Lindelöf, this gives you a very compact proof that $\Bbb R_\ell\times\Bbb R_\ell$ is not Lindelöf. If it were, its closed subspace $L$ would also be Lindelöf, but that is clearly not the case: no uncountable discrete space is Lindelöf, since the open cover by singletons has no countable subcover.
Your argument that $L$ is closed in $\Bbb R_\ell\times\Bbb R_\ell$ could use just a little more detail. Specifically, you should explain why the half planes whose union is $(\Bbb R_\ell\times\Bbb R_\ell)\setminus L$ are open in $\Bbb R_\ell\times\Bbb R_\ell$. This is easy enough to do: the lower-limit topology on $\Bbb R$ is finer than the Euclidean topology, so the topology on $\Bbb R_\ell\times\Bbb R_\ell$ is finer than the Euclidean topology on $\Bbb R^2$, and those half planes are open in the Euclidean topology, so they must be open in $\Bbb R_\ell\times\Bbb R_\ell$ as well.
You could have used a simpler open cover of $\Bbb R_\ell\times\Bbb R_\ell$: just start with the open cover
$$\mathscr{U}=\{(\Bbb R_\ell\times\Bbb R_\ell)\setminus L\}\cup\{U_x:x\in\Bbb R\}\,,$$
where $U_x=[x,x+1)\times[-x,-x+1)$. Then it is very clear that $$U_x\cap L=\{\langle x,-x\rangle\}$$ for each $x\in\Bbb R$ and hence that $U_x$ is the only member of $\mathscr{U}$ that contains $\langle x,-x\rangle$. It immediately follows that any subcover of $\mathscr{U}$ must include every one of the sets $U_x$, and you can simply say so. You actually ran into notational difficulty at this point when you tried to express a similar idea symbolically and wrote

$L \cap \{[-x,a) \times [x,b)|x,a,b \in \mathbb{R}, x<a,x<b\}=\{(-x,x)\}$.

This is simply false: elements of $L$ are points in $\Bbb R_\ell\times\Bbb R_\ell$, while elements of the the set on the righthand side of the intersection are subsets of $\Bbb R_\ell\times\Bbb R_\ell$, so in fact the intersection is empty. What you mean is that
$$L\cap\bigcup\big\{[-x,a)\times[x,b):x,a,b\in\Bbb{R},x<a,x<b\big\}=\{\langle -x,x\rangle\}\,.$$
By putting into $\mathscr{U}$ only one set containing any given point of $L$ I made it both a little easier to see that every one of those sets had to be kept in any subcover and definitely easier talk about what’s going on without getting bogged down in notation.
By the way, the open cover $\mathscr{U}$ that I used is called an irreducible open cover of $\Bbb R_\ell\times\Bbb R_\ell$: if you remove any one of its members, you no longer have a cover of the space.
