$\mathbb{R}$ 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: $\mathcal{B}_\ell$ 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 $\varnothing$ otherwise, so it is a member of $\mathcal{B}_\ell$ or empty. By [L1] the family is a basis for exactly one topology, and a set is open in $\mathbb{R}_\ell$ 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 $\mathcal{B}_\ell$, 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 $\mathbb{R}_\ell$ is not compact.

For claim 3 let $\mathcal{A}$ be an open cover of $\mathbb{R}_\ell$ and let $\mathcal{D}$ be the family of members of $\mathcal{B}_\ell$ 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 $\mathbb{R}_\ell \times \mathbb{R}_\ell$ the antidiagonal $\Delta$ is discrete as a subspace: for $x \in \mathbb{R}$ the basic set $[x, x+1) \times [-x, -x+1)$ meets $\Delta$ 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 $\Delta$ is therefore open in the subspace.

$\Delta$ is closed in $\mathbb{R}_\ell \times \mathbb{R}_\ell$: let $(u,v)$ have $u + v \ne 0$. 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 $\Delta$. If $u+v < 0$, put $\delta := -(u+v)/2 > 0$; 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 $\Delta$. So the complement of $\Delta$ is open.

$\mathbb{R} \setminus C$ is at most countable. Fix a surjection $\mathbb{N} \to \mathbb{Q}$ ([L4]) and for $x \notin C$ 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 $\mathbb{R}_\ell$ is Lindelöf: claim 3.

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