Improvement after J-M Schlenker's comment below :  

*This post has been divided into two parts, the second part is [here](https://math.stackexchange.com/questions/477438/uniquely-geodesic-spaces).*



> 
> **Question** : Is a finite dimensional metric space, [uniquely geodesic](http://en.wikipedia.org/wiki/Glossary_of_Riemannian_and_metric_geometry#U) if and only if it is [CAT(0)](http://en.wikipedia.org/wiki/CAT%28k%29_space) ?
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**In the case of a negative answer** :  

- Is *CAT(0)* assumption necessary ? Is it sufficient ?  

- What are the classical counter-examples ?  

- Is there a slight additive assumption for having a positive answer ?



Uniquely geodesic and CAT(0) spaces?
The CAT(0) asumption is sufficient: it implies that any two points are connected by a unique geodesic segment. This is well-known and follows from the definitions.


However, as pointed out by HenrikRüping, the CAT(0) asumption is not necessary, you can for instance perturb the hyperbolic plane by putting a small lump of positive curvature near a point and the resulting metric will not be CAT(0) but will still be uniquely geodesic.


You might however get a positive result in this direction if you add some smoothness and a (fairly strong) topological asumption, namely, that your space is the universal cover of a torus. Then you can use the result by Burago and Ivanov [here](http://link.springer.com/article/10.1007%2FBF01896241) that any Riemannian metric without conjugate point on a torus is flat. If a metric is uniquely geodesic then it is very close to having no conjugate points, and by their result it has to be flat, hence CAT(0). So as far as I can see it's conceivable that a uniquely geodesic distance on the universal cover of an $n$-dimensional torus is flat, hence CAT(0) (?)


