I need to prove that every suslin line is hereditarily Lindelof. Any idea will be helpfull.
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 nonempty 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), 664667
.
They prove in particular that a linearly ordered topological space $X$ satisfies the CCC if and only if $X$ is hereditarily Lindelof.

$\begingroup$ In fact $hL(X)=c(X)$ for all monotonically normal spaces. $\endgroup$ – Henno Brandsma May 29 '19 at 22:41