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I think this may be the case, and I've written a proof for it, but I'm not sure how rigorous it is.

Proof: Let S be a countable set, enumerated such that $$\forall x \in S, S=\{x_i\}_{i=1}^{\infty}.$$ Then, by the well-ordered principle, if $$\exists x=min(S),$$ then S can be ordered to be monotnically increasing such trhat $$x_i \le x_{i+1} \forall x \in S.$$

This would imply that if a set has a minimum element then it can be ordered from least to greatest; and similarly would provide structure for proof that if a set has a maximum element then it can be ordered from greatest to least.

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This is not correct. Consider the set $S$ of all nonnegative rational numbers. This set is countable, since the set of rational numbers is countable. And it has a minimum element, namely $0$. But you can't list out the rationals in increasing order like $x_1

It's hard to point out the flaw in your reasoning because it's hard to understand exactly what your reasoning is to begin with. The well-ordering principle is a property of the ordering of the natural numbers. Here you are considering some different set $S$ with a different ordering, so the well-ordering principle just doesn't tell you anything about it. The well-ordering principle says that any nonempty set of natural numbers has a least element, but it doesn't mean that just because some other ordered set happens to have a least element, then that set must "look like" the natural numbers.

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    Wouldn't non-negative rationals be well ordered with respect to the order induced from $\mathbb{N}$? though that order will be quite different from standard ordering of rationals – 2017-02-15
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    @user160738: Well, the phrasing of the question was a bit unclear about what ordering they were talking about. I assumed they were taking $S$ to already have an ordering (and they didn't want to change the order): otherwise, what does it mean to say that $\min(S)$ exists? I think when OP says "$S$ can be ordered to be monotonically increasing" they are not literally talking about defining a new order relation but are rather using "order" to mean "enumerate", as in "list out in some order". – 2017-02-15
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    @Eric Wofsey: Perhaps a sequence would work better for this, but I meant that if I have a set with elements that, if there is a minimum, I can rearrange the elements in the set to be in increasing order. I see why it doesn't hold for the set of rationals itself but it could work for subsets of the set of real numbers. – 2017-02-15
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    @PeterClark: Well, it will work for _some_ countable ordered sets. But it won't always work, as my example shows. I don't know exactly what more you're asking for. – 2017-02-15
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    @EricWofsey with the rationals, however, there is no smallest rational number, so you can't find min(Q) and enumerate it. – 2017-02-16
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    @PeterClark: My example is the nonnegative rationals, not the rationals. There is a smallest nonnegative rational. – 2017-02-16
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It looks like you're using a stronger assumption than you stated: you seem to use "every (nonempty) subset of $S$ has a minimal element."

This does in fact imply that $S$ is well-ordered (unlike the nonnegative rationals); however, it still doesn't help prove your claim! This is because, when you build your listing by induction, you have no way to argue that it contains every element of $S$; all you can conclude is that you've monotonically listed some of $S$.

For an explicit example of this, consider the set $$S=\{1-{1\over 2^n}: n\in\mathbb{N}\}\cup\{3-{1\over 2^n}: n\in\mathbb{N}\}.$$ It's not hard to show that every nonempty subset of $S$ has a least element; but the elements of $S$ can't be listed in increasing order. If you try your inductive construction, you get the listing $${1\over 2}, {3\over 4}, {7\over 8}, . . .$$ but e.g. ${5\over 2}$ never shows up here, even though it's in $S$.