A subset $S\in M$ is called strongly convex, or geodesically convex, if for any $p,q\in S$ there is a unique normal minimal geodesic $\gamma$ joining $p$ to $q$, and $\gamma$ is contained in $S$.
For example, on $(\mathbb{S}^{2},g_{\mathbb{S}^{2}})$, any geodesic ball of radius $r<\frac{\pi}{2}$ is strongly convex, (since, as I see, I take the unit sphere $\mathbb{S}^{1}$ on $\mathbb{R}^{2}$ his parametrization is $(\cos(\theta),\sin(\theta))$ then for any $r<\frac{\pi}{2}$ I will be still in the sphere, and as the sphere in $\mathbb{R}^{3}$ is the "same" the sphere $\mathbb{S}^{1}$ we're done).
And the book's claim, each strongly convex set is contractible, and the intersection of a family of strongly convex sets in $M$ is again strongly convex if it is not empty, I don't think is so trivial see this. And finally, can you give me an example of geodesically convex that does not belong to space, Thanks!!
The intersection rule is trivial. For any two points $p$ and $q$ in the intersection, they are both in each set, and therefore the unique geodesic $\gamma$ between them must have been in each set. But if the geodesic is in each set it is in the intersection. Therefore the definition is satisfied.
Contractibility seems harder. Let $p$ be a point in $M$. For $q$ in $M$ and $t$ in $[0, 1]$ define $f(t, q)$ to be the point a portion $t$ of the way along the unique geodesic $\gamma$ from $q$ to $p$. This is well-defined. Also $f(0, q) = q$ and $f(1, q) = p$. That has to be the contraction. The question is whether this map is continuous.
The challenge is this. If you have a triangle between two "nearby" points and a distant one, need the edges remain close? Well, in a general manifold, no. The "unique geodesic" condition has to come in somehow. But I can't see how to do it.
As for the last part of your question, the hyperbolic plane does not fit in space and is strongly convex.