In his Naïve Set Theory, Halmos defines, in Section 17, Well Ordering, the following:
We shall say that a well ordered set $A$ is a continuation of a well ordered set $B$, if, in the first place, $B$ is a subset of $A$, if, in fact, $B$ is an initial segment of $A$, and if, finally, the ordering of the elements in $B$ is the same as their ordering in $A$.
In Section 14, Order, he defines initial segments as following:
If $X$ is a partially ordered set, and if $a\in X$, the set $\{x\in X : x < a\}$ is the initial segment determined by $a$; we shall usually denote if by $s(a)$.
Question: If I understand correctly, then the second condition means that $B$ is the set of all strict predecessors of some $x$ in $A$. Then doesn't this condition imply the first one, namely that $B\subseteq A$?
For anyone else who might get confused (as I did) by Henno's answer below, the initial segments are not just sets, but are partially ordered sets having the partial order inherited from the parent set. (At least that's what Henno assumes.)
If we just know $A \subseteq B$, the orders on $A$ and $B$ could be unrelated, so we strengthen it by saying that $B=s(a)$ for some $a \in A$ which also tells us implicitly that $x < y$ means the same whether we are in $B$ or the larger $A$. So it does imply it but the first sentence "sets the scene" for your mind, as it were. It's not meant to be irredudant.