Why is a finite surface on a sphere cannot be covered by a single Cartesian coordinate? The infinitesimal line element on the surface of a sphere is given by $$ds^2=r^2(d\theta)^2+r^2\sin^2\theta(d\phi)^2\tag{1}$$ in spherical polar coordinates where the metric tensor is coordinate dependent. However, transforming to Cartesian coordinates, the same line element on the surface of the sphere is given by $$ds^2=(dx)^2+(dy)^2+(dz)^2\tag{2}$$ with $x^2+y^2+z^2=r^2$ so that the metric tensor becomes Euclidean i.e., $g_{ij}=\delta_{ij}$. 
I can intuitively understand that such a transformation lead to local Euclidean coordinates on the sphere but not global. How can I mathematically show that a single Cartesian frame cannot cover a finite region of the  surface of a sphere (or if we try to do so we get non-diagonal entries in the metric tensor)? 
I'm a student of physics, and my knowledge of mathematical jargon is limited. So please use minimum jargon if possible. 
 A: The two expressions for the metric on the sphere you've given are fundamentally different, in a sense more important than the local/global distinction - in fact, the Cartesian description is genuinely global while the spherical one has technical issues at the poles and at $\phi = 0,2\pi$ . The real difference is that $\theta,\phi$ are genuine coordinates on the two-dimensional sphere, while $x,y,z$ are coordinates on the three-dimensional space in which it is embedded. Any choice of $\theta,\phi$ corresponds to a point on the sphere, while only very particular combinations of $x,y,z$ do.
If "local Euclidean coordinates" means coordinates in which $g_{ij} = \delta_{ij}$, there are in fact no local Euclidean coordinates on the sphere, no matter how small you make your domain - any coordinate patch in which the expression for the metric is constant must necessarily be flat, but the sphere is everywhere curved. When we say a general surface/manifold is "locally Euclidean", it's at most in the topological sense that we can locally describe it by smooth coordinates (like $\theta,\phi$ in your example) - this puts no constraint on the geometry.
Thus there are three possible answers to your question, depending on how you fix it up: either we can't have a global coordinate system because this would make the sphere diffeomorphic to $\mathbb R^2$ (which is not true), we can't have a flat coordinate system because the sphere isn't flat (Gauss' Theorema Egregium), or actually, we can have global "Cartesian coordinates", so long as we remember the constraint.
