On continuously uniquely geodesic space

Question

This question was inspired by this comment of @68316.

Definition : A continuously uniquely geodesic space is a uniquely geodesic space whose geodesics vary continuously with endpoints.

Question : Is there a uniquely geodesic space which is not continuously uniquely geodesic ?

Answer 1

I got this example in the book, (exercise) example of a complete unique geodesic space which is not locally compact. Complete unique geodesic space but not proper

Answer 2

Update: This example is wrong, please check below comment from @Anton Petrunin

One example is the following, but there might be simpler ones:

Consider the set $$X = \{(r\cos t,r \sin t) \in \mathbb R^2 | 0 \leq r \leq 1,\ 0 < t \leq \pi/4\} \cup \{(x,0) | -1 \leq x \leq 0\} \cup \{(1,0)\}$$ equipped with the intrinsic metric induced from the standard metric of $\mathbb R^2$, i.e. the distance between points is the infimum of lengths of curves connecting them and lying in $X$. This is not a uniquely geodesic space since there is no geodesic connecting points $(x,0)$ and $(1,0)$ for $-1 \leq x \leq 0$. But if one identifies the points $(-1,0)$ and $(1,0)$ the quotient space, equipped with the quotient metic, is a uniquely geodesic space, say $Y$.

However $Y$ is not continous: Consider the sequence $v_n = (\cos \frac 1 n, \sin \frac 1 n)$. Then $v_n \to (1,0)$ but the geodesics $\gamma_n$ connecting $(0,0)$ and $v_n$ do not converge to the geodesic connecting $(0,0)$ and $(1,0)$, which is given by the segment $t \mapsto (-t,0)$, $t \in [0,1]$.

It is worth mentioning that $Y$ is not complete as a metric space. Finding complete examples is probably harder, if there are any. My guess would be that a complete uniquely geodesic space is in fact continous.