Is it possible for a non-well ordered but linearly ordered set to have a proper $<$-inductive subset? Say we have a set $S$, which is linearly ordered, but not well ordered. Let $B$ be a subset of $A$ such that $\forall t\in A$, $$\text{seg }t\subseteq B\implies t\in B$$
Here, $\text{seg }t=\{x\in A|x<t\}$. 
Is it possible that $B\neq A$? In other words, is it possible that $B$ is a proper subset of $A$?  
Motivation: My book says that if $B=A$, then $A$ is well-ordered. Hence, if $A$ is not well-ordered, it should be possible that $B\neq  A$. However, I cannot see how this would be possible, except if for every $t\in A$, $\text{seg }t\not\subseteq B$. Moreover, this is only possible if $A$ does not have a least element, as $\emptyset\subseteq B$, and $\emptyset$ is $\text{seg }t_0$, where $t_0$ is the least element of $A$. Hence, if $A$ has least element $t_0$, then $t_0\in B$,  and every subsequent  element in $A$ will fall into $B$.
Thanks.
 A: Sure.  Consider the $>$ relation on $\mathbb{N}$, and let $B$ be $\varnothing$.  Trivially for each $n \in \mathbb{N}$ we have that $\mathrm{seg}_> (n) = \{ m \in \mathbb{N} : m > n \} \nsubseteq \varnothing$, and so the implication $\mathrm{seg}_> (n) \subseteq \varnothing \rightarrow n \in \varnothing$ is true.
But clearly $\varnothing \neq \mathbb{N}$.

This can be generalised.  Suppose that $\langle X , < \rangle$ is any linearly ordered set which is not well-ordered.  Since it is not well-ordered, there is a strictly decreasing sequence $\langle x_n \rangle_{n \in \mathbb{N}}$ in $X$ (i.e., $x_{n+1} < x_n$ for all $n \in \mathbb{N}$).  Then $$B = \{ x \in X : ( \forall n \in \mathbb{N} ) ( x < x_n ) \}$$ can be shown to be an "inductive" subset which is not all of $X$.  (It may happen that $B = \varnothing$, but this is not always going to be the case.)
A: A non-trivial example: $B = [0..√2)∩ ℚ$ in $A = [0..∞) ∩ ℚ$ with the usual order. $B$ is closed in $A$ (in the order topology), so $∀ t ∈ A\colon\; \operatorname{seg} t ⊂ B → \overline{\operatorname{seg} t} ⊂ B$. This implies the condition required. Here, $A$ even has a least element.
The thing is, any set which does not contain any segments (or left-sided rays) in any linearly ordered set trivially satisfies the condition. So for example any bounded interval in $ℝ$ gives you such an example. Or any subset of $ℕ$ not containing $1$.
And this brings me to my question, as already stated in a comment: What do you mean by “if $B = A$, then $A$ is well-ordered”? Because, as I understand it, neither “if only $A$ has this property, then $A$ is well-ordered” nor “if $A$ (among maybe other subsets) has this property, then $A$ is well-ordered” is true.
