a totally ordered set with small well ordered set has to be small? doing something quite different the following question came to me:
1)If you have a totally ordered set A such that all the well ordered subset are at most countable, is it true that A has at most the cardinality of continuos?
2)More in general is it true that if a set A totally ordered has well ordered subset of lenght at most $|B|$ then A has at most cardinality $2^{|B|}$?

If $\kappa>2^\omega$, then $\kappa^*$ is a counterexample: all of its well-ordered subsets are finite. (The star indicates the reverse order.) However, if you require all well-ordered and reverse well-ordered subsets to be countable, the answer is yes to the more general question.
Let $\langle A,\le\rangle$ be a linear order with $|A|>2^\kappa$. Let $\preceq$ be any well-order on $A$. Then $[A]^2$, the set of $2$-element subsets of $A$, can be partitioned into sets $I_0$ and $I_1$, where 
$$I_0=\left\{\{a,b\}\in[A]^2:\le\text{ and }\preceq\text{ agree on }\{a,b\}\right\}$$
and
$$I_1=\left\{\{a,b\}\in[A]^2:\le\text{ and }\preceq\text{ disagree on }\{a,b\}\right\}\;.$$
By the Erdős-Rado theorem there are an $H\subseteq A$ and an $i\in\{0,1\}$ such that $|H|>\kappa$ and $[H]^2\subseteq I_i$. If $i=0$, $H$ is well-ordered by $\le$, and if $i=1$, $H$ is inversely well-ordered by $\le$.
(Nice question, by the way.)
