a totally ordered set with small well ordered set has to be small?

Solution 1:

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.)