well ordered and totally ordered set Can anyone please help me in this?

Prove that every well ordered set is a totally ordered set.

To start this proof will this be sufficient:
Let $x_1, x_2 \in X.$
Let $A=\{ x_1, x_2 \}$.
If $x_1$ is the least element, then $x_1\le x2$. If $x_2$ is the least element, then $x_1\ge x_2$.
How do I end the proof?
 A: Here is one possible definition and consequences:
Let $S$ be a totally-ordered set. $S$ is well-ordered if for every nonempty subset of $S$, there is a least element.
From the definition, $S$ is well-ordered $\Rightarrow$ there is a total-order on $S$ is trivial.
However, the converse is not necessarily true. For example, consider the reals, for which there is the usual total-order. Take any subset $A=\left(a,b\right)$. Clearly, this subset has no least element ($a \notin A$ and for every $p\in A$ there exists $q\in A$ with $q<p$).
A: To end the proof:
Thus either $x_1\le x_2$ or $x_2 \le x_1$.
As this holds for all $x_1,x_2\in X$, and as we have (perhaps?) assumed that $\le$ is an ordering, $(X,\le)$ is a totally ordered set.
A: It sounds like your definition of a well-order is as follows (and what we will use in this answer):

A set is well-ordered if and only if every nonempty subset $S$ has a least and minimal element: exactly one element $x$ such that $x \le y$ and $y \not < x$ for all $y \in S$.

We take $x < y$ to mean $x \le y$ and $x \ne y$.
Your approach is good.  You have prove linearity.  Now we need to prove that $\le $ is a partial order.
To prove that $x \le x$, set $S = \{ x \}$.  Then we only have one choice for a least element, so $x \le x$.
Now suppose $x \le y$ and $y \le x$.  Then set $S = \{ x,y \}$ and assume $x \ne y$.  Either $x$ or $y$ is a minimal element, so suppose $x$ is a minimal element (the same argument will work for $y$).  Then $y \not < x$ so $x = y$.
Now finally, let $x \le y$ and $y \le z$.  Set $S = \{ x,y,z \}$.  If $y$ is a least element then $x = y$, in which case $x \le z$ as desired.  If $x$ is a least element, then $x \le z$ trivially.  And if $z$ is a least element, then $z = y$ and $x \le z$, as desired.
