Prove that the given metric satisfies Einstein's equations in the vacuum For the Schwarschild metric
$$\mathrm{d}s^2=-\bigg(1-\frac{2GM}{r}\bigg) \mathrm{d}t^2 + \bigg(1-\frac{2GM}{r}\bigg)^{-1}\mathrm{d}r^2+r^2\mathrm{d}\theta^2 + r^2 \sin^2\theta \mathrm{d} \phi^2$$
prove that $\mathrm{d}s^2$ satisfies Einstein's equations in the vacuum, with $R_{\mu \nu} = 0$
Einstein's equations are:
$$R_{\mu \nu}-\frac{1}{2}Rg_{\mu\nu}+\Lambda g_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu \nu}$$
Since it is in the vacuum
$$T_{\mu \nu} = 0$$
Therefore the equation reduces to
$$\bigg(-\frac{1}{2}R + \Lambda\bigg)g_{\mu \nu} = 0$$
From the metric we see that
$$\mathrm{d}s^2=-\bigg(1-\frac{2GM}{r}\bigg) \mathrm{d}t \otimes\mathrm{d}t + \bigg(1-\frac{2GM}{r}\bigg)^{-1}\mathrm{d}r \otimes \mathrm{d}r +r^2\mathrm{d}\theta \otimes \mathrm{d}\theta + r^2\sin^2\theta \mathrm{d} \phi \otimes \mathrm{d} \phi$$
Therefore the metric tensor is
$$g=\begin{pmatrix}-(1-\frac{2GM}{r}) & 0 & 0 & 0 \\
0 & (1-\frac{2GM}{r})^{-1} & 0 & 0 \\
0 & 0 & r^2 & 0 \\ 0 & 0 & 0 & r^2 \sin^2\theta\end{pmatrix}$$
And the scalar curvature is obtained by
$$R=g^{ab}\bigg(\frac{\partial \Gamma^c_{ab}}{\partial x^c} - \frac{\partial \Gamma^c_{ac}}{\partial} + \Gamma^b_{ab}\Gamma^c_{bd} - \Gamma^d_{ac}\Gamma^c_{bd}\bigg)$$
But since, from the metric we can see that $g^{ab} = (g_{ab})^{-1}=0 \Rightarrow R=0$
furthermore $g_{\mu \nu} = 0$ for $\mu \neq \nu$
so the equation is satisfied for differend subindices.
After here I am unsure.
We are left with 4 equations
$$-\bigg(1-\frac{2GM}{r}\bigg) = 0$$
$$\bigg(1 - \frac{2GM}{r} \bigg)^{-1} = 0$$
$$r^2 = 0$$
$$r^2 \sin^2 \theta = 0$$
Equation (2) gives
$$\bigg(1 - \frac{2GM}{r}\bigg)^{-1} = \frac{r}{r-2GM}=0$$
$$r(r-2GM) = 0 \Rightarrow r = \{0, 2GM\}$$
This matches with the other 3 equations since, from equation (1) we can see that $r = 2GM$ and from equations 3 and 4 $r = 0$.
 A: The Einstein equation reduces to $R_{\mu \nu}=0$ when $T_{\mu \nu}=0$. To see this, set first $T_{\mu \nu}=0$ in the Einsten equation
\begin{equation}
R_{\mu \nu}-\frac{1}{2} R g_{\mu\nu}+\Lambda g_{\mu\nu} = 0
\end{equation}
then multiply it by $g^{\mu\nu}$ to get
\begin{equation}
g^{\mu\nu} R_{\mu \nu}-\frac{1}{2}g^{\mu\nu} R g_{\mu\nu}+g^{\mu\nu}\Lambda g_{\mu\nu} = 0
\end{equation}
Now $R= g^{\mu\nu} R_{\mu \nu}$ then the above equation becomes
\begin{equation}
R=\frac{D \Lambda}{\frac{D}{2}-1}
\end{equation}
where $D=g^{\mu\nu} g_{\mu\nu}$ is the spacetime dimension. In the last expression I used the Einstein summation convention. Then the vacuum equation in the case of non-zero cosmological constant can be written as
\begin{equation}
R_{\mu \nu } = \frac{\Lambda}{\frac{D}{2} - 1} g_{\mu \nu}
\end{equation} 
All off diagonal terms vanish. Now for $\mu \neq \nu$ you have 4 equations. I calculated $R_{\mu \nu}$ for a generic metric
\begin{equation}
\mathrm{d}s^2=-U(r) \mathrm{d}t^2 + V(r) \mathrm{d}r^2+r^2\mathrm{d}\theta^2 + r^2 \sin^2\theta \mathrm{d} \phi^2
\end{equation}
And I found
\begin{equation}
R_{00}=\frac{U_r}{r V_r}-\frac{U_r^2}{4 U V}-\frac{U_r V_r}{4 V_r^2}+\frac{U_{rr}}{2 V}
\end{equation}
\begin{equation}
R_{11}=\frac{U_r^2}{4 U_r^2}+\frac{V_r}{r V}+\frac{U_r V_r}{4 U V}-\frac{U_{rr}}{2 U}
\end{equation}
\begin{equation}
R_{22}=1+\frac{r V_r}{2 V^2}-\frac{r U_r}{2 U V}-\frac{1}{V}
\end{equation}
\begin{equation}
R_{33}=\sin(\theta)^2 R_{22}
\end{equation}
And the Ricci scalar is
\begin{equation}
R=\frac{2}{r^2}(1-\frac{1}{V})-\frac{U_{rr}}{U V}-\frac{2 U_r}{r U V}+\frac{U_r V_r}{2 U V^2}+\frac{2 V_r}{r V^2}+\frac{U_r^2}{2 U^2 V}
\end{equation}
Now I can compute the Einstein equations including the cosmological constant
\begin{equation}
R_{00}-\frac{1}{2} R g_{00}+\Lambda g_{00} =\frac{U V_r}{r V^2}+\frac{U}{r^2}-\frac{U}{r^2 V}-\Lambda U
\end{equation}
\begin{equation}
R_{11}-\frac{1}{2} R g_{11}+\Lambda g_{11} =\frac{U_r}{r U}+\frac{1}{r}-\frac{V}{r^2}+\Lambda V
\end{equation}
\begin{equation}
R_{22}-\frac{1}{2} R g_{22}+\Lambda g_{22} =\frac{r^2 U_{rr}}{2 U V}+\frac{r U_r}{2 U V}-\frac{r^2 U_r^2}{4 U^2 V}+r^2 \Lambda-\frac{r V_r}{2 V^2}-\frac{r^2 U_r V_r}{4 U V^2}
\end{equation}
\begin{equation}
R_{33}-\frac{1}{2} R g_{33}+\Lambda g_{33} =\sin(\theta)^2 (R_{22}-\frac{1}{2} R g_{22}+\Lambda g_{22})
\end{equation}
The above equation is not independent of the previous one. Assuming $U(r) \neq 0$, the first equation depends on $V(r)$ only, and it has solution
\begin{equation}
V(r)= \frac{1}{1+\frac{C}{r}-\frac{r^2 \Lambda}{3}}
\end{equation} 
where $C$ is the constant of integration. Once $V(r)$ is known we can compute $U(r)$ by using the second Einstein equation.
\begin{equation}
U(r)= (1+\frac{C}{r}-\frac{r^2 \Lambda}{3}) H
\end{equation}
where $H$ is the constant of integration. Then if $\Lambda \neq 0$ I have an additional term in the metric. The value of the constant can be determined by using physical considerations. For instance we expect that in the limit
that the mass tends to zero we should again obtain the flat metric. We see that this would result when $C=0$ so the intergration constant it is proportional to mass. I am not very familiar with the GR, but I think the metric you wrote satisfy the Einsten equation when $\Lambda$ is zero. 
There is another thing that it is worth to consider. According to your computation $r={0,2 G M}$, but this is not possible. The radius cannot be zero (this is a singularity on the metric). Concerning the value $2 G M$ I dont think the the Schwarzschild solution above does  hold, since it is a solution to the vacuum field equations.For instance if you consider the Sun $2 G M/c² = 2.95 km$ which is indide the Sun. On the other hand if a star collapse, its radius can be smaller than their Schwarzschild radius. In that case the surface is in the ‘vacuum’, outside the star, and the problems related to the Schwarzschild surface become real. However, this singularity is more like a
coordinate singularity than an actual singularity in the geometry. This means that a coordinate change can fix the mathematical problem since there is no physical singularity. 
