I am looking for a two Riemannian manifolds which are diffeomorphic, have the same constant curvature but they are not isometric.
I thought about the product manifold $(-\frac{\pi}{2}, \frac{\pi}{2})^n$ and $\mathbb{R}^n$, with the diffeomorphism: $f:(-\frac{\pi}{2}, \frac{\pi}{2})^n \rightarrow \mathbb{R}^n$ given by $f(x_1,\dots,x_n)=(\tan(x_1), \dots, \tan(x_n))$. Both manifolds have seccional curvature zero, since they are flat manifolds, but they are not isometric ($(-\frac{\pi}{2}, \frac{\pi}{2})^n$ is not complete).
Is that example right? Anyway, is there any "more intuitive" example?
To answer your question, this example is right, and your proof that it is right is right.
But I would say that there is a simpler and more intuitive proof that this example is right: the diameter of $(-\pi/2,\pi/2)^n$ is finite, whereas the diameter of $\mathbb R^n$ is infinite, and "finite diameter" is an isometry invariant.
Of course the diameter of a Riemannian manifold $M$ must be formalized, and this is done by defining $$\text{diam}(M) = \sup_{x,y \in M} d(x,y) $$ where $d(x,y)$ is the infimum of the lengths of all paths having endpoints $x$ and $y$. This formula yields a value $\text{diam}(M) \in [0,\infty) \cup \{\infty\}$ which is an invariant of Riemannian isometries.