Background
I have been doing some reading over winter break so far and found the idea that $\mathbb{R}$ has a well-ordering strange. So I have been thinking about it a little and was wondering if my proof for the following property is sound. If not, is it true? Proof critics are always encouraged! Thanks!
Statement
First, let me define a term
Definition: An ordering $\leq$ has a corresponding metric $d(x,y)$ if:
$$x < y < z \Rightarrow d(x,y) < d(x,z)\ \land\ d(y,z) < d(x,z)$$
The property I have tried to prove is
Theorem: Any well-ordering of $\mathbb{R}$ has no corresponding metric
Proof
Because of the well-ordered property, we can define $L_n$ to be the set containing the $n$ least elements in $\mathbb{R}$. $\forall x \in \mathbb{R}\setminus L_2$, there exists an $r_1 \in \mathbb{R}$ such that the open ball $B_{r_1}(x)$ has two unique least elements not equal to $x$. Call the smaller of the two $l$.
Because $B_{r_1}(x)$ is open, there exists some $\epsilon_1 > 0$ such that $B_{\epsilon_1}(l) \subset B_{r_1}(x)$.
Now define $r_2 := d(x, l)$. By definition of open ball, $l \not\in B_{r_2}(x)$. Because of well-order, $B_{r_2}(x)$ has some least element $l'$. There exists some $\epsilon_2 > 0$ such that $B_{\epsilon_2}(l') \subset B_{r_2}(x)$. So we have $l < l'$, $l \not\in B_{\epsilon_2}(l')$ and so $\forall y \in B_{\epsilon_2}(l)$, $y < l' \Rightarrow y \not\in B_{r_2}(x)$. Thus, $B_{\epsilon_2}(l) \cap B_{r_2}(x) = \emptyset$.
Now, let $\epsilon := \min(\epsilon_1, \epsilon_2)$. $B_\epsilon(l) = \{l\}$. It contains no elements less than $l$, as otherwise $l$ is not the min element of $B_r(x)$; it has no elements larger than $l$, as otherwise they would be in $B_{r_2}(x)$.
Thus, for any open ball $B$ with min element $l$, there exists $\epsilon > 0$ such that $B_\epsilon(l) = \{l\}$. Let $G$ be a set of all minimum elements of open balls. There is a clear injection from $G$ to $\mathbb{Q}$ by selecting any rational number in a small enough ball (using the regular metric) centered around elements of $G$.
Now notice that every open bounded interval is an open ball, and for any $x \in \mathbb{R}$, there exists some open interval $(x, b)$ where $b > x$. The smallest element in $(x, b)$ is the next element above $x$. Thus, for every $x \in \mathbb{R}$, the preceeding element of $x$ is in $G$ (and is unique with respect to $x$). This implies that since $\mathbb{R}$ is uncountable, so is $G$. This is our contradiction, and so a corresponding metric $d$ can not exist. QED.
Best Answer
Let $r^*$ be the minimum element in the well-order. For each real number $r$, let $d_r=d(r^*,r)\in [0,\infty)$. If $r^*<r<s$ (in the well-order), then $d_r<d_s$ (in the usual order), so $\{d_r\mid r\in \mathbb{R}\}$ is an uncountable subset of $[0,\infty)$ which is well-ordered (in the usual order). But this is impossible, because we can pick a rational number in each interval between $d_r$ and its successor.