Intrinsic metric with no geodesics It seems that I have the needed example, but I want it to be simple and self-explaining...

Construct a nontrivial complete metric space $X$ with intrinsic metric which has no nontrivial minimizing geodesics.

Definitions:


*

*A metric $d$ is called intrinsic if for any two points $x$, $y$ and any $\epsilon>0$ there is an $\epsilon$-midpoint $z$; i.e. $d(x,z),d(z,y)<\tfrac12 d(x,y)+\epsilon$. 

*A minimizing geodesic is nontrivial if it connects two distinct points.

*A meric space is nontrivial if it contains two distinct points.
Comments:


*

*Clearly, $X$ can not be locally compact.

 A: There is a very simple example of an intrinsic, complete metric space that is not geodesic (read in Ballmann's "Lectures on Spaces of Nonpositive Curvature": it is the graph on two vertices $x,y$, linked by edges $e_n$ of length $1+1/n$.
Of course it does not answer your question, but it may be possible to improve this example to one that does. Call $X_1$ the graph described above, and define $X_{n+1}$ from $X_n$
as follows: $X_n$ has a vertex $x'$ for each vertex $x$ of $X_n$, plus a vertex $v_e$ for each edge $e$ of $X_n$. For each edge $e=(xy)$ of $X_n$ we define edges $f_e^n$ and $g_e^n$ of $X{n+1}$: $f_e^n$ connects $x'$ to $v_e$ and has length $(1+1/n)$ times the original length of $e$, and $g_e^n$ does the same
but replacing $x'$ by $y'$.
Now it should be possible to construct the desired example by a limiting process. For example, take all vertices along the construction: the distance between any two of this points is constant as long it is defined, so we get a metric space. Its completion might be what you want (but I a not so sure of that after witting these lines).
A: There are metric simplicial graphs (each edge has length $1$) even quasi-isometric to the real line $\mathbb{R}$ (and as such Gromov hyperbolic) with no infinite geodesics: Start with the integers $\mathbb{Z}$ and connect any two distinct integers $x, y$ with a simplicial interval of length $|x-y|+1$ and otherwise disjoint from $\mathbb{Z}$. Any infinite geodesic must pass through the concatenation of two such intervals, but no concatenation is a geodesic -- it can be shortcut by a single interval. I hope this helps with the question you have in mind.
A: This does not answer the question. The following paper contains a very natural example, namely the regular Frechet Lie group $Diff_{\mathcal S}(\mathbb R)$ of all diffeomorphisms of the real line which fall rapidly towards the identity, with a weak right invariant Riemannian metric induced by inner product
$$
G_{Id}(X,Y) = \int_{\mathbb R} X'Y'\,dx
$$
on the Lie algebra, which has the following property: 


*

*Geodesic distance is an intrinsic metric, but there does not exist a single nontrivial geodesic. But it is not complete as a metric space.


This group is a normal subgroup (isometrically contained for the geodesic distance, by thm 4.5) in the slightly larger regular Lie group 
$Diff_{\mathcal S_1}(\mathbb R)$ where one allows shifts at $+\infty$. 
The inner product extends to the larger space of vector fields in the Lie algebra. The resulting weak right invariant Riemannian metric  (see 4.3) is flat,
has minimal geodesics between any two points (it is geodesically convex), it allows for a formula for geodesic distance, and it has a nice geodesic completion which is a monoid.
The geodesic equation is the Hunter-Saxton PDE in the real line. 
Any geodesic in  $Diff_{\mathcal S_1}(\mathbb R)$ hits the subgroup $Diff_{\mathcal S}(\mathbb R)$ at most twice.
Geodesic distance on $Diff_{\mathcal S}(\mathbb R)$ is an intrinsic metric; this follows from the proof of theorem 4.5.


*

*Martin Bauer, Martins Bruveris, Peter W. Michor: Homogeneous Sobolev metric of order one on diffeomorphism groups on the real line. Journal of Nonlinear Science 24, 5 (2014), 769-808 (pdf)
A: Well, the unit ball in $c_0$ is almost what you want (there is no unique shortest curve between points). All we need now is to enhance "bypasses" and to give disadvantage to "straight lines". This can easily be done by taking the distance element to be $(2+\sum_n 2^{-n}x_n)^{-1}\|dx\|_\infty$, which is never less than the usual distance element in $c_0$ and never greater than 3 times it in the unit ball. Now, if we have any continuous finite length curve $x(t)$ from $y$ to $z$ parametrized by the arclength, we can easily shorten it by replacing the $m$-th position by the maximum of the actual value of $x_m(t)$ and $y_m+t(z_m-y_m)/d+\frac 12 \min(t,d-t)$, where $d$ is the length of $x(t)$, which will work if $m$ is large enough since $\max_t|x_m(t)|\to 0$ as $m\to\infty$ and both functions change slower than the distance along the original curve.
This is certainly self-explaining (the shortest curve escapes from $c_0$ to $\ell^\infty$) but I do not know if it is simple enough for your purposes.  
