why is there no order in metric spaces? Why is there no abstract notion of order in the metric spaces? 
Surely, if there is a notion of distance there must be a notion of different values. If there is a notion of different values, why can't we specify a well defined order in the metric space to reflect the lowest to greatest values in the space?
 A: There need not be a meaningful such relation on a metric space. The basic reason you omit it is because it by omitting requirements cover up more cases. When talking about metric space we restrict the requirements to only mean that there exists a distance between the elements and nothing more (we do not require that the space has an origin, we do not require that the elements can be added or scaled and so on).
For example apart from the normal cases $\mathbb R$ and $\mathbb C$, the concept covers a lot of different cases. For example the trivial metric space with only one element, hamming metrics (the number of symbols that differ between two strings), euclid geometrics (that doesn't rely on the existence of an origin) and so on.
If you try to define the order naively by size you will run into the fact that a metric space doesn't have the notion of size either, it's defined in terms of distance between elements. And even if there were (and then it's not merely a metric space, but maybe a normed vector space) you would end up with something that does not fulfill the requirements of an order (you don't have the property that if $a\le b$ and $b\le a$ then $a=b$).
If you want an ordered metric space, then you should require that and not hope for that all abstract "structures" should conform to your requirements. 
Math consists of a zoo of abstract "structures" from the most generic ones to more specialized - you have to choose which one to use in every occation (but if you use more generic one, your conclusions will become more generic).
A: We can certainly derive a metric from certain notions of size (norms), but need not have any notion of size at all to have a metric.
Let $X$ be any set, and for $x,y\in X,$ define $$d(x,y)=\begin{cases}1 & x\ne y\\0 & x=0.\end{cases}$$ This is a metric on $X$ that tells us when two elements are different, but nothing else.
A: Order in itself has little to do with distance. I can order a set of words by alphabetical order. What matters is the order relationship: transitive, asymmetric, etc.
The reals both have an order and a metric. That makes them awesome. I could impose an order on the unit square by applying a space filling curve to the unit interval. It would be like unraveling a sweater. This order would have nothing to do with distance in $R2$, or anything else that makes two dimensions interesting. The sweater, once unraveled, is no longer a sweater. 
A: The set of complex numbers $\Bbb{C}$ is clearly a metric space under the usual notion of modulus, i.e., $(\Bbb{C},d)$ is a metric space where $d(w,z)=|w-z|=\sqrt{(w-z)(w-z)^{\ast}}$. However, at the same time, $\Bbb{C}$ has no total ordering relation that is 'meaningful'. This is especially intuitive if we think of $\Bbb{C}$ as a plane analogous to $\Bbb{R}^{2}$. This is a concrete example of a metric space which has no (meaningful) ordering relation.
As of now, I have failed to prove to myself that there does not exist an order that can be constructed on $\Bbb{C}$ using only the metric. But my point is that, if such an order exists and is a total order, then it is 'meaningless'. Hence, considering orders constructed just from a metric might be an interesting problem in some specific instances, but may not be a great use of time in the general case, as some metric spaces are important for other, incompatible, reasons.
A: Consider any possible ordering on the points on a circle. You will find that:


*

*it is not invariant under isometries;

*some of the sets $\{x: x<a\}$
are not open.
This tells you that the ordering cannot have so much relation with the metric structure...
A: I think this question suggests an answer addressing whether or not every metric topology is also induced by a linear order. The answer is no. I'll try to sketch a proof of this by means of an example.
Consider the topological space given by three copies of the closed unit interval where we identify the zeros. This is a metric space. Namely let $I_1=[0,1]\times\{1\}$ and $I_i=(0,1]\times\{i\}$ for $i\in \{2,3\}$ and the metric $d$ on $X=\cup_i I_i$ be given by $d((x,i),(y,i))=|x-y|$ and $d((x,i),(y,j))=|x+y|$ and $d((0,1),(y,j))=y$ for any $x,y\in (0,1]$ and any $i\neq j$. See Hedgehog space.   
Suppose that the topology $\tau$ induced by $d$ is also generated by some linear order $<$ on $X$. We denote the intervals in $<$ with endpoints $x$ and $y$ by $(x,y)_<$. Consider the point $x_0=(0,1)$. Note that there must be two distinct $i,j\in\{1,2,3\}$ such that either $(x_0,y)_{<} \cap I_i \neq \emptyset$ and $(x_0,y)_{<} \cap I_j \neq \emptyset$ for any $y>x_0$ or $(y,x_0)_{<} \cap I_i \neq \emptyset$ and $(y,x_0)_{<} \cap I_j \neq \emptyset$ for any $y<x_0$. Without loss of generality we consider the first case. Let $y_i>x_0$ be a point in $I_i$ and then let $x_0<y_j<y_i$ be a point in $I_j$. The open intervals $(-\infty, y_j)$ and $(y_j,+\infty)$ partition $\{x_0\}\cup I_i$ into two open subspaces. However $\{x_0\}\cup I_i$ is homeomorphic to $[0,1]$ and hence connected. Contradiction. 
It is worth noting that the converse is also not true. Namely, not all topologies induced by a linear order and metrizable. For example the space $[0,\omega_1]$, where $\omega_1$ denotes the first uncountable ordinal, with the topology induced by ordinal order, is compact but not separable, so it is not a metric space. 
A: Let's consider the set of all points on the surface of the Earth and measure the distance between two points by the length of the shortest path over the surface. This is a metric space.
How are the points ordered? Which point is the greatest and which is the least?
A: As other answers pointed out, the answer to your question is negative. But there is a very close natural construction. Let $(A, d_A)$ be a metric space. The set $A\times\mathbb{R}$ can be endowed with a natural product metric. Define the relation $\le$ on the metric space $A\times\mathbb{R}$ by $(a, \lambda)\le(b, \mu)$ if and only if $$d_A(a,b)\le\mu-\lambda.$$ It is not difficult to show that $(A\times\mathbb{R}, \le)$ is a partial ordered set.
According to this construction, if you can decompose a given metric space $(X, d_X)$ such that $X=A\times\mathbb{R}$ and $d_X=d_A+\vert\_\vert$, where $(A, d_A)$ is another metric space, then you have an obvious partial ordering on $X.$
