Is proper time equal to the Invariant Interval or the time elapsed in the Rest Frame? In SR consider two time like separated events -
In some frame \begin{equation}ds^2= dt^2 - dx^2\end{equation}
In a frame where the events occur at the same place ( rest frame; $dx' =0$) then according to what I know proper time is the time elapsed in that frame i.e. $d\tau=dt'$.
Hence \begin{equation}ds'^2 = d\tau^2 =dt'^2\end{equation} ( since $d\tau =dt'$ in that frame)and since the interval is invariant;
\begin{equation}
d\tau^2=dt'^2= dt^2 - dx^2.\end{equation}
Consider the same event in GR, in some frame ( coordinate system) \begin{equation}ds^2= g_{00}dt^2 - g_{11}dx^2....\end{equation}
In the frame ( coordinate system) where the events occur at the same place ( $dx'^2=0$) then according to what I know proper time is the time elapsed in that frame i.e. $d\tau=dt'$,  and we should have, \begin{equation}
ds'^2= g_{00}dt'^2 = g_{00}d\tau^2\end{equation} ( by the same analogy as in SR, $d\tau =dt'$) and since the interval is invariant, we should have, \begin{equation}
g_{00}d\tau^2 = g_{00}dt^2 - g_{11}dx^2....\end{equation}.
But from time dilation formula in GR, I know this wrong.
Precisely, according to what I have unsertood, proper time is the time elapsed in the rest frame of the particle, just like in SR, so for the interval\begin{equation}
ds^2= g_{00}dt^2 - g_{11}dx^2....\end{equation}, if two events happen at the same place
\begin{equation}ds^2 = g_{00}dt'=g_{00}d\tau^2\end{equation} ( by definition just as in the SR case).

*

*Why is this wrong. Is my reasoning that proper time is the time measured in the rest frame wrong
                         Or 

2)is that the time coordinate in a general metric not represent time measured by any clock.
 A: This is not a contradiction, but simply a constraint on the allowable coordinate systems that would qualify as a particle’s rest frame. You have discovered that along the worldline of the particle the metric in the particle’s rest frame must have $g_{00}=1$. It is not mandatory to use such coordinates, but only such coordinates will be called the particle’s rest frame.
As an example, consider the standard Schwarzschild metric for a manifold with (-+++) signature and in units where $c=1$ $$g=\left(
\begin{array}{cccc}
  -\left(1-\frac{R}{r}\right) & 0 & 0 & 0 \\
 0 & \left(1-\frac{R}{r}\right)^{-1} & 0 & 0 \\
 0 & 0 & r^2 & 0 \\
 0 & 0 & 0 & r^2 \sin (\theta ) \\
\end{array}
\right)$$ which is the line element $$ds^2 = -d\tau^2 = -\left(1-\frac{R}{r}\right) dt^2 + \left(1-\frac{R}{r}\right)^{-1} dr^2 + r^2 d\theta^2 + r^2 \sin (\theta ) d\phi^2$$
Now, for a particle at rest we have $dr=d\theta=d\phi=0$ so $$ds^2 = -d\tau^2 = -\left(1-\frac{R}{r}\right) dt^2 = g_{00} dt^2$$ Notice that for $r=\infty$ we have $g_{00}=-1$ so the proper time $d\tau$ is equal to the coordinate time $dt$. So these coordinates are a valid rest frame for an observer at rest at $r=\infty$. However, for $r=10R$ we have $ g_{00} = -0.9$ so $d\tau^2 = 0.9 dt^2$ thus the coordinate time is not equal to the proper time. These same coordinates are not a valid rest frame for an observer at rest at any finite $r$.
A: Your second statement is correct. In GR, one can always change to a new set of coordinates $x'=x'(x)$, so it's clear that any particular choice of time coordinate doesn't have any absolute physical significance. What is invariant is the interval $s$. An observer moving along some timelike path will experience time elapsing according to the invariant interval along the path, and this is independent of the choice of coordinates. Now, if you are that observer, one thing you could do is watch your clock, and "draw a tick mark" on the spacetime once per second according to your clock. By doing this you would have constructed a time coordinate which has the simple property that $ds = dt$. But this relation only holds for that choice of time coordinate.
Edit to follow up on some comments:
You are wondering whether/why in SR $g_{00}$ must be equal to $-1$. The answer is that it doesn't matter. Or more precisely, that's a meaningless statement. The reason is that time is a dimensionful quantity. If I measure time in seconds, and I have $g_{00}=-1$, then I could also keep the same time coordinate but set $g_{00}=-1/{3600}$ and then I would get the time measured in minutes. So it's really a moot point. The difference with GR is that in GR, the metric varies over spacetime. So the worldline of an observer can pass through points where the metric is different, and those relative changes will produce interesting effects. In GR, you would make some choice of units at one point on your worldline, and you could use that to set $g_{00}=-1$ at that point, but then as you continue on the worldline, you would experience different values of the metric, and thus the value of coordinate time and your accumulated invariant interval will start to differ.
