Why aren't the lines in taxicab geometry the geodesics of the $L^1$ metric?

Geodesics in the Taxicab plane are not unique. A trivial example of this is given by the points $(0, 0)$ and $(1, 1)$ and the curves formed by going from $(0, 0)$ to $(0, 1)$ then $(1, 1)$ and the other as going from $(0, 0)$ to $(1, 0)$ then $(1, 1)$. The two segments forming the curves are easily seen to both have length 1, and the sum total of those lengths is then 2, so these curves both have length 2. But the distance between $(0, 0)$ and $(1, 1)$ is $|1 - 0| + |1 - 0| = 2$ so both are geodesics. In general, any "stair-step" which always moves parallel to either the $x$- or $y$-axes with no backtracking is a geodesic as can be seen by translating its component segments to match these simpler geodesics. Such a path is a continuous-space version of a Dyck path as described here:

http://mathworld.wolfram.com/DyckPath.html

Thus if we define a line as a geodesic it is subject to a great deal of ambiguity, in particular there are $\beth_1$ geodesics between any two points (it can't be more since geodesics must be continuous). In particular, there would be tremendous ambiguity in the "betweenness" relation as you define it -- I'd believe that in fact any point lying in the rectangle with points $A$ and $C$ would end up as being "between" in this definition, in the sense that "there would exist a line (geodesic) from $A$ to $C$ that passes through the given point".

Yup, it would: Let $A = (x_A, y_A)$, $C = (x_C, y-C)$, and $B = (x_B, y_B)$ with $x_A < x_B < x_C$ and $y_A < y_B < y_C$. Then

$$d(A, B) = |x_B - x_A| + |y_B - y_A|$$ $$d(B, C) = |x_C - x_B| + |y_C - y_B|$$

so

$$ \begin{align} d(A, B) + d(B, C) &= |x_B - x_A| + |y_B - y_A| + |x_C - x_B| + |y_C - y_B|\\ &= (x_B - x_A) + (y_B - y_A) + (x_C - x_B) + (y_C - y_B)\\ &= x_B - x_A + y_B - y_A + x_C - x_B + y_C - y_B\\ &= x_C - x_A + y_C - y_A\\ &= (x_C - x_A) + (y_C - y_A)\\ &= |x_C - x_A| + |y_C - y_A|\\ &= d(A, C) \end{align} $$

So yeah, betweenness this way works terribly. Unless perhaps you want to argue that Taxicab geometry should be a "geometry of solid rectangles" ... but then how is it any different from Euclidean geometry of rectangles I wonder?

(FWIW the midset of Taxicab geometry, that is, the "perpendicular bisector" is FREAKY!!! Scared and disturbed me when I first saw it! I can still feel that fweeky tingley little heebiejeebies feeling a little coming on thinking about it reading right now I'm also getting that with what I just proved above, there is something conceptually horrifying about the idea that my brain makes out that this is like a line that is like, eeeevshz, infinitely swollen or something...)

So we have to narrow things down. It turns out though that Euclidean lines are also a geodesic. The most general definition of arc length of a curve to a metric space $M$ given as $C: [0, 1] \rightarrow M$ is

$$L = \sup \sum_{i=1}^{N} d(C(t_{i+1}), C(t_i))$$

where the supremum is over all partitions $0 = t_1 < t_2 < \cdots < t_N = 1$ of $[0, 1]$. So let $C(t) = (t, kt)$ be a Euclidean line in Taxicab space that is not vertical (vertical can be dispensed with a change of coordinates). Then $d(C(t_{i+1}), C(t_i)) = |t_{i+1} - t_i| + k|t_{i+1} - t_i| = t_{i+1} - t_i + k(t_{i+1} - t_i) = (1 + k)(t_{i+1} - t_i)$ and

$$ \begin{align} L &= \sup \sum_{i=1}^{N} (1 + k) (t_{i+1} - t_i)\\ &= \sup (1 + k) \sum_{i=1}^{N} t_{i+1} - t_i\\ &= \sup (1 + k) (1)\\ &= \sup (1 + k)\\ &= 1 + k \end{align} $$

which is easily seen to be the distance between the start and end points. Note that we used telescoping sum in the above. This can be visualized as the "fallacious" method used to get "$\pi = 4$" in the following answer:

The staircase paradox, or why $\pi\ne4$

This method is correct for Taxicab geometry, and the image correctly shows that the ratio of the circumference of a Euclidean circle to its diameter in taxicab geometry is 4.

Thus, to get a sensible geometry, we pick the Euclidean line as the choice geodesic for the taxicab system.