$\mathbb{R}$ with the half-open intervals $[a,b)$ as a basis is not compact and, assuming the Axiom of Countable Choice, is Lindel"of, while its square is not Lindel"of, the antidiagonal being an uncountable closed discrete subspace

Claim 1: ${\mathcal{B}}_{\mathrm{\ell }}$ covers $\mathbb{R}$, since $x\in [x,x+1)$; and $[a,b)\cap [c,d)$ is $[\max \{a,c\},\min \{b,d\})$ when that is nonempty and $\mathrm{\varnothing }$ otherwise, so it is a member of ${\mathcal{B}}_{\mathrm{\ell }}$ or empty. By [L1] the family is a basis for exactly one topology, and a set is open in ${\mathbb{R}}_{\mathrm{\ell }}$ exactly when each of its points has a half-open interval around it inside it.

Claim 2: the family $\{[-\iota (n),\iota (n)):n\in \mathbb{N},n\ge 1\}$ consists of members of ${\mathcal{B}}_{\mathrm{\ell }}$, hence of open sets, and covers $\mathbb{R}$ by [L3]; the members increase with $n$, so a finite subfamily has union $[-\iota (N),\iota (N))$ for the largest index $N$ occurring, which omits $\iota (N)$. So ${\mathbb{R}}_{\mathrm{\ell }}$ is not compact.

For claim 3 let $\mathcal{A}$ be an open cover of ${\mathbb{R}}_{\mathrm{\ell }}$ and let $\mathcal{D}$ be the family of members of ${\mathcal{B}}_{\mathrm{\ell }}$ contained in some member of $\mathcal{A}$; by step 1.1 the family $\mathcal{D}$ covers $\mathbb{R}$. Put $C:=\bigcup \{(a,b):[a,b)\in \mathcal{D}\}$.

In ${\mathbb{R}}_{\mathrm{\ell }}\times {\mathbb{R}}_{\mathrm{\ell }}$ the antidiagonal $\mathrm{\Delta }$ is discrete as a subspace: for $x\in \mathbb{R}$ the basic set $[x,x+1)\times [-x,-x+1)$ meets $\mathrm{\Delta }$ only in $(x,-x)$, since a point $(y,-y)$ in it satisfies $x\le y$ and $-x\le -y$, that is $y\le x$. By [L7] each singleton of $\mathrm{\Delta }$ is therefore open in the subspace.

$\mathrm{\Delta }$ is closed in ${\mathbb{R}}_{\mathrm{\ell }}\times {\mathbb{R}}_{\mathrm{\ell }}$: let $(u,v)$ have $u+v\mathrm{\ne }0$. If $u+v>0$, every point $(y_1,y_2)$ of $[u,u+1)\times [v,v+1)$ has $y_1+y_2\ge u+v>0$, so the box misses $\mathrm{\Delta }$. If $u+v<0$, put $\delta :=-(u+v)\mathrm{/}2>0$; every point of $[u,u+\delta )\times [v,v+\delta )$ has $y_1+y_2$ at least $u+v$ and less than $u+v+2\delta =0$, so again the box misses $\mathrm{\Delta }$. So the complement of $\mathrm{\Delta }$ is open.

$\mathbb{R}\setminus C$ is at most countable. Fix a surjection $\mathbb{N}\to \mathbb{Q}$ ([L4]) and for $x\mathrm{\notin }C$ let $r(x)$ be the rational of least index with $x<r(x)$ and $[x,r(x))\in \mathcal{D}$; such rationals exist, since $\mathcal{D}$ covers gives $[a,b)\in \mathcal{D}$ with $a\le x<b$, a rational $q$ with $x<q<b$ by [L3] then has $[x,q)\subseteq [a,b)$ and so $[x,q)\in \mathcal{D}$. Nothing is selected, the least index being determined by $x$. The map $r$ is injective on $\mathbb{R}\setminus C$: if $x<y$ lay outside $C$ with $r(x)=r(y)=q$, then $[x,q)\in \mathcal{D}$ gives $(x,q)\subseteq C$ and $x<y<q$ puts $y$ in $C$. So $\mathbb{R}\setminus C$ injects into $\mathbb{Q}$, hence is equinumerous with a subset of $\mathbb{Q}$ and at most countable by [L4].

$C$ is covered by the at most countable family ${\mathcal{D}}_{\mathbb{Q}}:=\{[p,q)\in \mathcal{D}:p,q\in \mathbb{Q}\}$, at most countable because $[p,q)\mapsto (p,q)$ injects it into $\mathbb{Q}\times \mathbb{Q}$, which is at most countable by [L4], as is therefore the image subset: given $x\in C$ there is $[a,b)\in \mathcal{D}$ with $a<x<b$, and [L3] gives rationals $p,q$ with $a<p<x<q<b$, whence $[p,q)\subseteq [a,b)$ lies in $\mathcal{D}$ and contains $x$.

So ${\mathcal{D}}_0:={\mathcal{D}}_{\mathbb{Q}}\cup \{[x,r(x)):x\in \mathbb{R}\setminus C\}$ is an at most countable subfamily of $\mathcal{D}$ by [L4] and covers $\mathbb{R}$ by steps 3.1 and 3.2. Every member of $\mathcal{D}$ lies inside some member of $\mathcal{A}$, so [L6] applied to an indexing of ${\mathcal{D}}_0$ by $\mathbb{N}$ supplies one member of $\mathcal{A}$ for each member of ${\mathcal{D}}_0$, and those form an at most countable subcover of $\mathcal{A}$. Hence ${\mathbb{R}}_{\mathrm{\ell }}$ is Lindelöf: claim 3.

$\mathrm{\Delta }$ is uncountable, being in bijection with $\mathbb{R}$ under $x\mapsto (x,-x)$ and $\mathbb{R}$ being uncountable by [L4]. Were ${\mathbb{R}}_{\mathrm{\ell }}\times {\mathbb{R}}_{\mathrm{\ell }}$ Lindelöf, its closed subspace $\mathrm{\Delta }$ would be too: given a cover of $\mathrm{\Delta }$ by traces of ambient open sets, adjoining the complement of $\mathrm{\Delta }$ gives an ambient open cover, an at most countable subcover of it traces back to an at most countable subcover of $\mathrm{\Delta }$. But $\mathrm{\Delta }$ is discrete by step 2.3, so its singletons form an open cover admitting only itself as a subcover, and that family is uncountable. So ${\mathbb{R}}_{\mathrm{\ell }}\times {\mathbb{R}}_{\mathrm{\ell }}$ is not Lindelöf: claim 4.