Deriving Birkhoff's Theorem
The Birkhoff's Theorem in 3+1D is e.g. proven (at a physics level of rigor) in Ref. 1 and Ref. 2. (An elegant equivalent 1-page proof of Birkhoff's theorem is given in Refs. 3-4.) Imagine that we have managed to argue$^1$ that the metric is of the form of eq. (5.38) in Ref. 1 or eq. (7.13) in Ref. 2:
$$ds^2~=~-e^{2\alpha(r,t)}dt^2 + e^{2\beta(r,t)}dr^2 +r^2 d\Omega^2. \tag{A} $$
It is a straightforward exercise to calculate the corresponding Ricci tensor $R_{\mu\nu}$, see eq. (5.41) in Ref. 1 or eq. (7.16) in Ref. 2. The notation is here $$x^0\equiv t, \quad x^1\equiv r, \quad x^2\equiv\theta, \quad\text{and} \quad x^3\equiv\phi.$$ The Einstein's equations in vacuum read
$$R_{\mu\nu}~=~\Lambda g_{\mu\nu}~.\tag{E} $$
The argument is now as follows.
From $$0~\stackrel{(E)}{=}~R_{tr}~=~\frac{2}{r}\partial_t\beta$$ follows that $\beta$ is independent of $t$.
From $$0~\stackrel{(A)}{=}~\Lambda\left(e^{2(\beta-\alpha)} g_{tt}+g_{rr} \right) ~\stackrel{(E)}{=}~ e^{2(\beta-\alpha)} R_{tt}+R_{rr}~=~\frac{2}{r}\partial_r(\alpha+\beta) $$ follows that $\partial_r(\alpha+\beta)=0$. In other words, the function $f(t):=\alpha+\beta $ is independent of $r$.
Define a new coordinate variable $T:=\int^t dt'~e^{f(t')}$. Then the metric $(A)$ becomes $$ds^2~=~-e^{-2\beta}dT^2 + e^{2\beta}dr^2 +r^2 d\Omega^2.\tag{B}$$
Rename the new coordinate variable $T\to t$. Then eq. $(B)$ corresponds to setting $\alpha=-\beta$ in eq. $(A)$.
From $$\Lambda r^2~\stackrel{(B)}{=}~\Lambda g_{\theta\theta} ~\stackrel{(E)}{=}~ R_{\theta\theta} ~=~1+e^{-2\beta}\left(r\partial_r(\beta-\alpha)-1\right) ~=~1-\partial_r(re^{-2\beta}), $$ it follows that $$ re^{-2\beta}~=~r-R-\frac{\Lambda}{3}r^3 $$ for some real integration constant $R$. In other words, we have derived the Schwarzschild-(anti)de Sitter solution, $$e^{2\alpha}~=~e^{-2\beta}~=~1-\frac{R}{r}-\frac{\Lambda}{3}r^2.$$
Finally, if we switch back to the original $t$ coordinate variable, the metric $(A)$ becomes
$$\begin{align}ds^2~=~&-\left(1-\frac{R}{r}-\frac{\Lambda}{3}r^2\right)e^{2f(t)}dt^2 \cr &+ \left(1-\frac{R}{r}-\frac{\Lambda}{3}r^2\right)^{-1}dr^2 +r^2 d\Omega^2.\end{align}\tag{C}$$
It is interesting that the metric $(C)$ is the most general metric of the form $(A)$ that satisfies Einstein's vacuum equations. The only freedom is the function $f=f(t)$, which reflects the freedom to reparametrize the $t$ coordinate variable.
References:
Sean Carroll, Spacetime and Geometry: An Introduction to General Relativity, 2003.
Sean Carroll, Lecture Notes on General Relativity, Chapter 7. The pdf file is available here.
Eric Poisson, A Relativist's Toolkit, 2004; Section 5.1.1.
Eric Poisson, An Advanced course in GR; Section 5.1.1.
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$^1$ Here we for convenience show how Ref. 1 and Ref. 2 reduce from
$$\begin{align} ds^2~=~&g_{aa}(a,r)~da^2 +2g_{ar}(a,r)~ da~dr \cr &+g_{rr}(a,r)~ dr^2 +r^2d\Omega^2\end{align} \tag{5.30/7.5}$$
to
$$ ds^2~=~m(r,t)~dt^2 +n(r,t)~dr^2 +r^2d\Omega^2. \tag{5.37/7.12}$$
Proof: Define a function
$$n~:=~g_{rr}-\frac{g_{ar}^2}{g_{aa}}$$
and an inexact differential
$$ \omega~:=~da+\frac{g_{ar}}{g_{aa}}dr.$$
Then eq. (5.30/7.5) reads
$$ ds^2~=~g_{aa}\omega^2 +n~dr^2 +r^2d\Omega^2.$$
The function $\sqrt{m}$ in eq. (5.37/7.12) can be viewed as an integrating factor to make the differential $\sqrt{\frac{g_{aa}}{m}}\omega$ exact, i.e. of the form $dt$ for some function $t(a,r)$.
Are you doing this rigorously by using an ansatz for your metric and plugging it into the usual Einstein field equations in vacuum or something "more topological"?
If former, in the most common ansatz,
$ds^2 = e^{f(t,r)} dt^2 - e^{g(t,r)} dr^2 - r^2 d\Omega^2 $
you get your metric independence of time t with the
{01}th Ricci tensor component, which sets a time derivative of one of your metric components to 0. Algebraic combinations of the other Ricci tensor componentes give you the relationships between metric component functions $f$ and $g$, somewhere along the way you should get something like $\frac{d}{dt}[f(t,r)-g(t,r)] = 0$. That gives you your time independance of $g_{11}$.