I thought this was easy until I realized that for a set to be well-ordered, EACH of it's non-empty subsets must have a least element.
Let $A$ be a non-empty totally ordered set. Let $x_0$ be some element of $A$. Define $S=\{x\in A|x\ge x_0\}$. $x_0 \in S$ so $S$ isn't empty.
Now, this seems like it would work. It certainly is totally ordered and cofinal in $A$, but it might not be well-ordered.
For example, if $A=\mathbb R$ then $(x_0 + 1, x_0 + 2)$ (where $()$ denotes an open interval) is a non-empty subset of $S$, but doesn't have a least element.
Best Answer
Let $\langle X,\le\rangle$ be a non-empty linear order. If $X$ has a maximum element $x_0$, then $\{x_0\}$ is a well-ordered cofinal subset, so assume that $X$ has no largest element. Let
$$\mathscr{W}=\{W\subseteq X:\langle W,\le\rangle\text{ is a well-order}\}\;,$$
and define a relation $\preceq$ on $\mathscr{W}$ by setting $W_0\preceq W_1$ if and only if $W_0$ is an initial segment of $W_1$.