For the lower-limit line, $\chi=d=L=c=\aleph_0$ and $w=2^{\aleph_0}$ under choice

Assume Choice and let $S$ be the lower-limit line. At $x$, the intervals $[x,x+1/n)$ form a countable local base. No finite family is a local base: the intersection of finitely many neighbourhoods is still a neighbourhood and contains some $y>x$, whereas $[x,y)$ cannot contain any member of a finite local base contained in that intersection. Hence $\chi(S)=\aleph_0$.

The rationals are countable and meet every nonempty half-open interval, so $d(S)\le\aleph_0$; no finite set is dense, since a short half-open interval can avoid it. The published lower-limit-line lemma gives Lindelöfness, while the cover $\{[-n,n):n\ge1\}$ has no finite subcover, so $d(S)=L(S)=\aleph_0$. Density bounds cellularity above, and the disjoint family $\{[n,n+1):n\in\mathbb Z\}$ bounds it below, giving $c(S)=\aleph_0$.

All half-open intervals form a basis of cardinality at most $|\mathbb R|^2=|\mathbb R|$. Conversely, well order any basis and assign to each $x$ its first member $B_x$ with $x\in B_x\subseteq[x,x+1)$; if $x<y$, then $B_y$ cannot contain $x$, so $B_x\ne B_y$. Thus $w(S)=|\mathbb R|=2^{\aleph_0}$.