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`](http://www.ams.org/journals/proc/1969-023-03/S0002-9939-1969-0248762-3/).