path metrics without geodesics This is a follow-up of this question. Recall that a metric space $(X,d)$ is called a path-metric space if the distance between any two points in $X$ equals the infimum of lengths of paths between these points. Such $(X,d)$ contains a plenty of rectifiable paths (unless $X$ is either empty or is a one-point set). A geodesic between the two points $x, y\in X$ is a 
path $c$ from $x$ to $y$, whose length equals the distance $d(x,y)$. 
Question. Is there a complete path-metric space $(X,d)$ (of infinite cardinality) such the only geodesics in $(X,d)$ are constant maps to $X$? (I think so, but do not see a clear proof.) 
Obviously, such space $X$ cannot be locally compact (at any point). 
Note that in the example given in the linked question, there are pairs of points without geodesics connecting them. Another example (an infinite-dimensional complete Hilbert manifold) is mentioned in this wikipedia article. However, in both examples, the space contains plenty of geodesics (Hilbert manifolds always do). 
Edit. I just discovered that the same question was posted (in 2010) and answered on Mathoverflow here. 
 A: Just an extended comment (initially an answer but I misread the question): here's an example that is not complete.
Consider the union $D$ of all circles in the Euclidean plane of rational center and nonzero rational radius, with the distance induced from $\mathbf{R}^2$.
1) This is a length (=path) metric space. Indeed, pick $x,y\in D$ and fix $\varepsilon>0$. Pick a little open piece of circle $X\subset D$, of radius $\le\varepsilon$, through $x$, and similarly $Y$ through $y$. The set of lines intersecting both $X$ and $Y$ has non-empty interior, hence contain a rational line; using this, we can find a rational line $E$ intersecting both $X$ and $Y$; it contains rational points at $x',y'$ at distance $\le\varepsilon$ of $x$ resp $y$. The bisector line of $[x',y']$ is rational, hence we can find pieces of circle with rational center far away on this bisector line, that are arbitrary close to the segment $[x',y']$, and eventually these circles do intersect $X$ and $Y$. This yields a path in $D$ from $x$ to $y$ with, say, length $\le d(x,y)+(1+\pi)\varepsilon$.
2) The space $D$ contains no segment. Indeed the intersection of each line with a circle consists of at most 2 points, and hence the intersection of each line with $D$ is countable.
A: In order to close the question: 
The same question was posted (in 2010) and answered on Mathoverflow here. 
