Is a totally ordered set well-ordered, provided that its countable subsets are? 
Let $(X,⪯$) be totally ordered set, prove that if every non empty countable subset of $X$ is well ordered then X is well ordered.

It does seem obvious that any subset should have a minimum but I am not sure of how to prove it. Any help would be appreciated, and sorry I put up the same problem earlier but forgot to include the 'countable' subset.
 A: If $X$ were not well-ordered, you'd have a nonempty $Y\subseteq X$ with no smallest element.  As $Y$ is nonempty, pick some $y_0\in Y$.  As $y_0$ isn't minimal in $Y$, pick some $y_1\prec y_0$ in $Y$. As $y_1$ isn't minimal in $Y$, pick some $y_2\prec y_1$ in $Y$. Continue in this fashion, obtaining an infinite sequence $y_0\succ y_1\succ\dots\succ y_n\succ y_{n+1}\succ\dots$.  The set $\{y_n:n\in\mathbb N\}$ is a countable subset of $Y$ with no smallest element.
A: If we assume that the Principle of Dependent Choices (see the fourth paragraph in particular) holds, then this is true. It needn't be true in Zermelo-Fraenkel set theory without the Axiom of Choice (abbreviated ZF).
Suppose that $X$ is not well-ordered, so that some non-empty subset $B$ of $Y$ without a $\preceq$-least element. That is, for each $x\in B$ there is some $y\in B$ such that $y\prec x$. Then by PDC, there is a sequence of points $x_1,x_2,x_3,...$ of $B$ such that $x_{n+1}\prec x_n$ for all $n$. But then $\{x_n:n=1,2,3...\}$ is a non-empty countable subset of $X$ without $\preceq$-least element, so it is not well-ordered by $\preceq$. By contrapositive, the proposition holds.
It's worth noting that the converse does hold in ZF. That is, if $X$ is well-ordered by $\preceq$, then so is every non-empty subset of $X$ (in fact, so is every subset of $X$).
A: Pick a non-empty subset of X, namely A (without loss of generality it's infinite, otherwise you find a minimum since every countable subset of X has it); if A doesn't have a minimum element then by the totally ordered hypotesis you get a countable strictly decreasing chain in A, but it must have a minimum (according to your hypothesis). 
