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If I am not wrong, the surface of a sphere minus a point is a metric vector space that is isomorphic to a non-euclidean $\mathbb{R}^2$ space. Thus in principle there is no way to find a system of coordinates in which the metric looks Euclidean. But, if you choose as coordinates $\phi$ and $\theta$ (assuming r=1), you can define the scalar distance $s$ as $s^2=\phi^2 +\theta^2$, which looks euclidean.

Is there any reason why this does not work as an euclidean distance? (I know I am wrong on something but I do not know what I am missing)

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    I'm not sure I understand the question, correct me if I'm wrong. The sphere is compact so no matter what metric you pick you will always have this property that the sphere achieves the maximum distance between any two points. Unlike $\mathbb{R}^2$. But this property also holds if you remove the point from a sphere. So I think you misunderstand one thing: a sphere without a point is homeomorphic to $\mathbb{R}^2$. It doesn't mean that the metrics are equal, actually there is no such homeomorphism preserving the distance. – 2017-02-27
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    In fact it's euclidean. You can define whatever you want, the problem will be how to fit all the things you are working with. With that metric, the geodesics are loxodromes: https://en.wikipedia.org/wiki/Rhumb_line – 2017-02-27
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    @freakish I do not know much about compactness, and I am not sure it is important to my question (I thought the "minus-a-point" was necessary, but if not a full sphere is ok). My question is 1) is the sphere and example of elliptic geometry? 2) but If this is so, how it is possible to define an euclidean distance? – 2017-02-27
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    @RafaBudría now I am confused, then my statement that for a sphere "there is no way to find a system of coordinates in which the metric looks Euclidean" is wrong? – 2017-02-27
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    Aside: you should note that the punctured sphere is not a vector space in any natural way. Hence when you say "isomorphic" you are being imprecise. If you view the punctured sphere as a metric subspace of $\Bbb{R}^3$, then it is not isometrically equivalent to $\Bbb{R}^3$ with the usual metric, but it is homeomorphic to it. Just saying "isomorphic" doesn't say which of these two possibilities you have in mind. – 2017-02-27
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    @RafaBudría would this mean that a being that lives in such "curved" space will believe that his space is "flat"? I believed that the geometry was defined only by the form of the metric – 2017-02-27
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    @RobArthan I guess I have to learn more about the subject, something seems wrong: if the sphere (with or without a point) is not a vector space, then why is it given as the poster child of non-euclidean geometry? ( I assumed the sphere was a vector space, otherwise how could it be an example of a metric vector space) – 2017-02-27
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    Yes, I meant that: that space is flat for a being living there. But consider this: from your point of view, their "straight lines" doesn't seem straight to you: they are loxodromes!! The problem you've heard of (a sphere has not a coordinate system for wich the sphere is flat) is not about definitions but about the metric the sphere embedded in the usual tridimensional space has, and the sphere isn't obviously flat!! it's clearly curved. The assumption is you are measuring distances with usual 3D rulers! but along the sphere's surface. – 2017-02-27
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    @HughMungus: the sphere supplies an example of a non-Euclidean geometry when viewed as a [Riemannian manifold](https://en.wikipedia.org/wiki/Riemannian_manifold). No vector space structure on the sphere is required or relevant for this. – 2017-02-27

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