How limiting/ heavy is the "triangle inequality" assumption? Suppose a theorem proves something about a family of distance measures, with this the triangle inequality assumption. How limiting this assumption is in reality?
What are some real-world examples of distance measures which don't obey the triangle inequality? 

Edit:
The triangle inequality is the following requirement upon the distance function on $M$:
$$
\operatorname{dist}(x,z) \leq \operatorname{dist}(x,y) + \operatorname{dist}(y,z), \qquad \forall x,y,z \in M.
$$
 A: That pretty much depends on your definition of distance "in the real world". In my opinion, the triangle inequality is extremely important for the definition of any notion of distance. 
The triangle inequality tells you that going from $A$ to $B$ and taking a detour over $C$ will at most increase the distance that needs to be covered to get from $A$ to $B$. 
Assuming that the triangle inequality does not hold would mean that the distance from $A$ to $C$ plus the distance from $C$ to $B$ is less than the distance from $A$ to $B$. But then nothing would stop you from declaring that smaller distance as the "real" distance from $A$ to $B$...
One thing to maybe keep in mind is that the mathematical notion of distance refers to the minimal effort one needs to make to go from one point to another.
A: A real-life "distance" with a weakened triangle inequality is the round-trip delay time.
EDIT: the RTT is the time it takes for a signal to make the journey back and forth between two nodes on a network. According to The Inframetric Model for the Internet:

A large amount of algorithms has recently been designed for the Internet under the assumption that the distance defined by the round-trip delay (RTT) is a metric.

But...

... we confirm that the triangle inequality does not hold for a significant fraction of the nodes. Nevertheless, we demonstrate that RTT measures
satisfy a weak version of the triangle inequality: there exists a small constant $\rho$ such that for any triple  $u$, $v$, $w$, we have $RTT(u,v)\le\rho\max\{RTT(u,w),RTT(w,v)\}$.

A: This doesn't answer the question you asked, but if you're interested in real world examples, a natural thing to do is to relax the assumption of symmetry. The result is called a quasimetric. Explicitly, this is a function $$d : X^2 \rightarrow \mathbb{R}_{\geq 0} \cup \{\infty\}$$ satisfying the following axioms:


*

*$d(x,x)=0$

*$d(x,y)=0 \rightarrow x=y$

*$d(x,y)+d(y,z) \geq d(x,z)$


(I'll talk about $\infty$ later.)
I wish these were called "difficulty" functions (as opposed to the academic-sound "quasimetric"), because they're are good at describing the difficulty of getting from one place to another. For example, imagine you're in a mountainous, infinite world. For concreteness, we take the underlying set to be $\mathbb{R}^2.$ We define $d(x,y)$ as the difficulty of getting from $x$ to $y$. Perhaps think of this as the amount of food rations you have to consume to get from $x$ to $y$. Our axioms then say:


*

*The amount of food rations needed to not move anywhere is $0$.

*If the number of food rations needed to get somewhere is $0$, then that somewhere is the same point you're already standing on.

*If the number of food rations needed to get from where you're standing to some point $y$ is $Y$ and the number of food rations to get from there to another point $z$ is $Z$, then the number of food rations needed to get from where you're standing to $z$ is at most $Y+Z$. (This makes sense because worst case scenario, you can walk the most efficient path from where you are to $y$, consuming $Y$ rations, and then from there walk the most efficient path to $z$, consuming a further $Z$ rations.)


Okay, so what about symmetry?
Well, suppose $x$ is at the bottom of a mountain that takes many days to climb, and $y$ is at the top. Then maybe it takes $10$ food rations get from $x$ to $y$, but only $5$ to get from $y$ to $x$. More colloquially: the difficulty of getting from one point to another needn't equal the difficulty of getting back again. So symmetry is definitely out.
By the way, the reason to include $\infty$ is to describe places that you can't get to, no matter how much effort you put in. For instance, imagine the world consists of two islands, and its too far for you to swim from one to the other, or back again. Then if $x$ is an element of island $0$ and $y$ is an element of island $1$, we'd expect: $$d(x,y) = \infty, \qquad d(y,x) = \infty.$$
Notice that you need both the above assumptions, because without symmetry, you cannot derive $d(x,y) =\infty$ from $d(y,x)=\infty$, nor vice versa. This is actually a feature, not a bug. Suppose you want to model a landscape that has a building in it. Furthermore, assume that if you're inside the building, you can press a button to exit; but if you're outside, there's no way to get in. This can be described by asserting that $d(x,y) = \infty$ whenever $y$ is an element of the building, while letting $d(y,x)$ be some finite value depending on $x$.
A: As @GenericNickname pointed out, this very much depends on what your definition of distance is. However, one interesting real world example is traveling on a curved surface, such as airplanes on earth.
If you look at a flap map projection, then you cannot use Euclidian geometry. This is because straight lines on a map are not necessarily straight on a globe. This is why airplane paths look curvy: it is actually the shortest distance from one point to another.
Also, I'm not sure how rigorous of an answer you are looking for; this is just an interesting case that came to mind.
A: Assuming the triangle inequality does not hold for an infinite number of points, then you can always find an H such that dist(A,B) < dist(A,H) + dist(H,B). This gives you a "shorter" path between A and B. But now you repeat this process with another point to further narrow down dist(A,H) and dist(H,B). In the end, you will probably end up with zero-length distances as optimal paths.
How useful can this be? Perhaps you aren't interested in the distances themselves, but how long it takes to get from A to B. In the real world, think of a GPS device that has to calculate how to go from A to B. Sometimes, A-B directly is the best route. Sometimes, depending on weather, traffic and other factors, going "around the triangle" is faster. 
Perhaps you can come up with a coordinate system that has shifting "weights" between paths (say, a vector field. Imagine mountains coming and going out of nowhere) that change over time, and so the "optimal" path changes, along with the validity of the triangle inequality.
