We are given two points, $(0,0)$ and $(1,1)$. How many shortest paths exist between these points?
For $p=2$ (Euclidean) distance, the answer is $1$, as the shortest path is a straight line. However, if we use $p=1$ (manhattan) distance, there are an infinite number of lines connecting the two points with length $2$. (E.g. the graph of any monotonically increasing continuous function $f$ with $f(0)=0$ and $f(1)=1$.)
Also, what is the envelope of these paths?
For $1<p<\infty$, strict convexity of the norm implies that the only shortest path is the line segment between these points.
For $p=1$, the length of a path $(x(t),y(t))$ is just the sum of the lengths of one-dimensional paths $x(t)$ and $y(t)$. Both of those must go from $0$ to $1$. They will have length $1$ if and only if the function is nondecreasing. Hence, the shortest paths are those with nondecreasing $x$ and $y$ coordinates. They will up the square in which $(0,0)$ and $(1,1)$ are opposing vertices.
To deal with $p=\infty$ in two dimensions, it's probably easiest to map it isometrically onto $\ell^1$ by the transformation $u=\frac{x+y}{2}$, $v=\frac{x-y}{2}$. Check that $|u|+|v| = \max(|x|,|y|)$... Then we look for the shortest paths from $(0,0)$ to $(1,0)$ in the $\ell^1$ norm. The argument from preceding paragraph applies: the coordinates $u,v$ must be nondecreasing functions of $t$. This means $v\equiv 0$ and $u$ is nondecreasing. Thus, a shortest path between these points must be the line segment.
