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An uncountable subset of the real numbers with the standard ordering ≦ cannot be a well order: Suppose X is a subset of R well ordered by ≦. For each x in X, let s(x) be the successor of x in ≦ ordering on X (unless x is the last element of X). Let A = { (x, s(x)) | x ∈ X } whose elements are nonempty and disjoint intervals. Each such interval contains at least one rational number, so there is an injective function from A to Q. There is an injection from X to A (except possibly for a last element of X which could be mapped to zero later). And it is well known that there is an injection from Q to the natural numbers (which could be chosen to avoid hitting zero). Thus there is an injection from X to the natural numbers which means that X is countable. On the other hand, a countably infinite subset of the reals may or may not be a well order with the standard "≦". For example,

・The natural numbers are a well order under the standard ordering ≦.
・The set {1/n : n =1,2,3,...} has no least element and is therefore not a well order under standard ordering ≦.
Examples of well orders:
・The set of numbers { - 2-n | 0 ≦ n < ω } has order type ω.
・The set of numbers { - 2-n - 2-m-n | 0 ≦ m,n < ω } has order type ω2. The previous set is the set of limit points within the set. Within the set of real numbers, either with the ordinary topology or the order topology, 0 is also a limit point of the set. It is also a limit point of the set of limit points.
・The set of numbers { - 2-n | 0 ≦ n < ω } ∪ { 1 } has order type ω + 1. With the order topology of this set, 1 is a limit point of the set. With the ordinary topology (or equivalently, the order topology) of the real numbers it is not.

References
1＾S. Feferman Some applications of the notions of forcing and generic sets Fundamenta Mathematicae (1964)
http://matwbn.icm.edu.pl/ksiazki/fm/fm56/fm56129.pdf

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