Question about proper time in general relativity I think I may have some fundamental misunderstanding about what $dt, dx$ are in general relativity. 
As I understand it, in special relativity, $ds^2=dt^2-dx^2$, we call this the length because it is a quantity that is invariant under Lorentz boosts. If a ball is moving in space and I want to calculate the $ds$ for the ball to travel from point A to point B, then $d\tau=ds$  (where $\tau$ is proper time) because according to the ball, $dx=0$, since point A and point B are at the same place in its own frame.
Assuming this is correct, this is all fine with me. Now fast forward to general relativity. $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$...so if a ball is travelling in spacetime under whatever metric, then to me it would seem like, in the ball's frame, we should set $dx^i=0$, and then we get $ds^2=g_{00}d\tau^2$...however, from what I am told and from what I have read, $ds^2=d\tau^2$. When I look at the Schwarzschild metric, for example, $g_{00}$ doesn't appear to be 1 in a travelling ball's frame.
What have I misunderstood about the interval $ds$?. Is it that $ds$ only purely has to do with geometry of spacetime, and doesn't quite represent the distance between events? 
 A: The problem arose when you wrote $ds^2 = g_{00} d\tau^2$. Generally one of your coordinates $x^\mu$ will be timelike, and the others spacelike, but the timelike one is not in general the proper time of someone whose spatial coordinates are not changing. That is, $t \neq \tau$. Using your sign convention,1 $d\tau^2 = ds^2$, so the (arbitrarily large) lapse in proper time between two events A and B is
$$ \Delta\tau = \int\limits_\text{path} \sqrt{ds^2} = \int_{\sigma_\mathrm{A}}^{\sigma_\mathrm{B}} \sqrt{g_{\mu\nu} \frac{\mathrm{d}x^\mu}{\mathrm{d}\sigma} \frac{\mathrm{d}x^\nu}{\mathrm{d}\sigma}} \ \mathrm{d}\sigma. $$
Here $\sigma$ is any parameter that parametrizes the specific path (required to be timelike) taken from A to B, and your location along the path at parameter $\sigma$ has coordinates $x^\mu(\sigma)$.
Suppose $x^i \equiv 0$ along the path. Also call $x^0$ by the name $t$. Then most of the terms in the sum vanish and we have
$$ \Delta\tau = \int_{t_\mathrm{A}}^{t_\mathrm{B}} \sqrt{g_{00}} \ \mathrm{d}t. $$
In fact we could have gotten this just by examining
$$ d\tau^2 = g_{\mu\nu} \mathrm{d}x^\mu \mathrm{d}x^\nu. $$
Now often in GR we do call that timelike coordinate $t$, as I have done. Usually when this is done, as in the Schwarzschild metric
$$ ds^2 = \left(1 - \frac{2M}{r}\right) \mathrm{d}t^2 - \left(1 - \frac{2M}{r}\right)^{-1} \mathrm{d}r^2 - r^2 \left(\mathrm{d}\theta^2 + \sin^2(\theta) \ \mathrm{d}\phi^2\right), $$
that $t$ becomes arbitrarily close to the $\tau$ of a local observer "at rest" in these coordinates as you move into the flat regime. In this case, as you move away from the mass at small $r$, $t \to \tau$. In other words, you can only neglect the $g_{00}$ term in some asymptotic cases (which you'll note correspond to $g_{00} \to 1$). In your SR example, you happened to use coordinates that matched those of the observer whose proper time you cared to measure, but you could have boosted to a new frame, in which case the $\mathrm{d}t^2$ part of the metric would have had some prefactor involving $\gamma$.

1 I will point out that most often in GR proper, $ds^2$ is negative for timelike separations, with $d\tau^2 = -ds^2$. The convention you adopted is more common in particle physics.
A: Expanded on Michael Brown's comments, you seem to have confused $dt$ with $d\tau$.  They are not the same.  It is true that $ds^2 = dt^2$, but also true that $ds^2 = g_{tt} dt^2 + \ldots$ (other terms omitted).
What this means physically is that, even for two events that take place at the same location but different coordinate times (according to a given frame), there is a difference between elapsed coordinate time and actual (or proper) elapsed time.
This can happen in special relativity, too.  Choose a time coordinate $\bar t$ such that $dt = \alpha \, d\bar t$, and the metric would look like $ds^2 = \alpha^2 \, d\bar t^2 = d\tau^2$.  What this should emphasize is that one's choice of time coordinate can end up being quite poor, not corresponding well (if at all!) to the physically meaningful quantity, proper time.
While it may be confusing for the time coordinate to not correspond exactly to proper time, it nevertheless gives an enormous amount of freedom to recast the metric in a way that is convenient--you do not have to have proper time be your time coordinate if you don't want it to be, and it will always have a well-defined expression regardless of your choice of coordinates, as all physical quantities must have.
