How to show the spacetime interval is invariant in general? I understand how to derive the spacetime interval being invariant for Minkowski space, but I've never seen any derivation of it in general curved spacetime. Is the invariance just derived for Minkowski space and then postulated that it holds for all metric tensors in general relativity, or is there a proof to show it is invariant in general relativity?
 A: You cannot derive the invariance of the line element because it is one of the assumptions on which relativity (both flavours) is based. When you say:

I understand how to derive the spacetime interval being invariant for minkowski space

I would guess you mean that you can show the Lorentz transformations preserve the line element. However most of us would take the view that the invariance of the line element was more fundamental, then derive the Lorentz transformations from the requirement that the line element be preserved.
There isn't a simple equivalent to the Lorentz transformations in general relativity. The Lorentz transformations are a coordinate transformation but a very simple one where the transformation is between inertial frames in flat spacetime. While we use coordinate transformations extensively in GR they are usually far more involved than the Lorentz transformations.
However in GR, just as in SR, the invariance of the line element:
$$ ds^2 = g_{\alpha\beta} dx^\alpha dx^\beta $$
always applies though the metric $g_{\alpha\beta}$ is generally more complicated.
A: Lets look at an arbitrary invertable coordinate transformation: $$ x^\mu \rightarrow x'^{\mu}=x'^{\mu}(x^\nu).  $$
The corresponding Jacobian $\Lambda$
$$ \Lambda^\mu_{~~\rho}=\frac{\partial x'^{\mu}}{\partial x^{\rho}}$$ is invertable 
$$ \Lambda_{\sigma}^{~~\nu}=\frac{\partial x^{\nu}}{\partial x'^{\sigma}}.$$
A vector tansforms like $$x'^\mu=\frac{\partial x'^{\mu}}{\partial x^{\sigma}}x^\sigma=\Lambda^\mu_{~~\sigma}x^\sigma.$$
The defining property of a tensor of second rank (the metric tensor is such a tensor)  is that it transforms like $$g'_{\rho\sigma}=\frac{\partial x^{\mu}}{\partial x'^{\rho}}\frac{\partial x^{\nu}}{\partial x'^{\sigma}}g_{\mu\nu}=\Lambda_{\rho}^{~~\mu}\Lambda_{\sigma}^{~~\nu}g_{\mu\nu}.$$
With that in mind lets give this tensor calculus a go on our line element:
\begin{align} ds'^2 & = g'_{\mu\nu}dx'^\mu dx'^\nu \\\\ & =g'_{\mu\nu}\Lambda^\mu_{~~\rho}\Lambda^\nu_{~~\sigma}dx^\rho dx^\sigma\\\\ & = g_{\mu\nu}dx^\rho dx^\sigma \\\\ & = ds^2.\end{align}
That would be the textbook calculation for the invariance of the line element using tensor calculus. To prove the transformation property of a second rank tensor one would express everything via base vectors and use the relations of those base vectors.
So the invariance of the line element is more a feature of tensor calcus. A scalar is invariant under coordinate transformations that has nothing to with special or general relativity.
A: The spacetime interval is a concept of Minkovski spacetime. It does also appear in general relativity in its infinitesimal form $ds$, as the principles of special relativity apply locally within curved spacetime of general relativity. In general relativity, the distances between two points in curved spacetime are described by geodesics or by a path integral over $ds$.
