This is not really a research problem. The answer is in exercise 3.12.4(b,e) in Engelking's General Topology text. For each linearly ordered topological space $X$ we have that the cellularity $c(X)$ of $X$ equals the hereditarily Lindelöf number of $X$ (and you ought to be able to prove this). By definition every Souslin line has countable cellularity, i.e. every disjoint family of non-empty open sets is at most countable. The hereditarily Lindelöf number, $hL(X)$, is defined as $\sup\{L(Y): Y \subseteq X\}$ where $L(Y)$ is the Lindelöf number of $Y$, i.e. every open cover of $Y$ has a subcover of cardinality at most $L(Y)$.

The proof that $hL(X)\ge c(X)$ is trivial (and holds for every space $X$).

The other direction, that $hL(X)\le c(X)$ for linearly ordered spaces, is a result of Lutzer and Bennett, Separability, the countable chain condition and the Lindelöf property in linearly orderable spaces, Proc. Amer. Math. Soc. 23 (1969), 664-667.

This is not really a research problem. The answer is in exercise 3.12.4(b,e) in Engelking's General Topology text. For each linearly ordered topological space $X$ we have that the cellularity $c(X)$ (also called Souslin number) of $X$ equals the hereditarily Lindelöf number of $X$ (try to prove it on your own). By definition every Souslin line has a countable cellularity (which is another term for the countable chain condition, CCC), i.e. every disjoint family of non-empty open sets is at most countable. The hereditarily Lindelöf number, $hL(X)$, is defined as $\sup\{L(Y): Y \subseteq X\}$ where $L(Y)$ is the Lindelöf number of $Y$, i.e. every open cover of $Y$ has a subcover of cardinality at most $L(Y)$.

The proof that $hL(X)\ge c(X)$ is trivial (and holds for every topological space $X$).

The other direction, that $hL(X)\le c(X)$ for linearly ordered topological spaces, is a result of Lutzer and Bennett, Separability, the countable chain condition and the Lindelöf property in linearly orderable spaces, Proc. Amer. Math. Soc. 23 (1969), 664-667. They prove in particular that a linearly ordered topological space $X$ satisfies the CCC if and only if $X$ is hereditarily Lindelof.

This is not really a research problem. The answer is in exercise 3.12.4(e) in Engelking's General topology text. For each linearly ordered topological space $X$ we have that the cellularity of $X$ equals the hereditarily Lindelof number of $X$ (and you ought to be able to prove this). By definition every Souslin line has countable cellularity, i.e. every disjoint family of non-empty open sets is at most countable. The hereditarily Lindelof number, $hL(X)$, is defined as $\sup\{L(Y): Y \subseteq X\}$ where $L(Y)$ is the Lindelof number of $Y$, i.e. every open cover of $Y$ has a subcover of cardinality at most $L(Y)$.

This is not really a research problem. The answer is in exercise 3.12.4(b,e) in Engelking's General Topology text. For each linearly ordered topological space $X$ we have that the cellularity $c(X)$ of $X$ equals the hereditarily Lindelöf number of $X$ (and you ought to be able to prove this). By definition every Souslin line has countable cellularity, i.e. every disjoint family of non-empty open sets is at most countable. The hereditarily Lindelöf number, $hL(X)$, is defined as $\sup\{L(Y): Y \subseteq X\}$ where $L(Y)$ is the Lindelöf number of $Y$, i.e. every open cover of $Y$ has a subcover of cardinality at most $L(Y)$.

The proof that $hL(X)\ge c(X)$ is trivial (and holds for every space $X$).

The other direction, that $hL(X)\le c(X)$ for linearly ordered spaces, is a result of Lutzer and Bennett, Separability, the countable chain condition and the Lindelöf property in linearly orderable spaces, Proc. Amer. Math. Soc. 23 (1969), 664-667.