A brute force (and ugly) derivation of the Killing fields of Schwarzschild metric
The Schwarzschild metric is
\begin{equation}
ds^2 = -\left(1-\frac{R_{\text{S}}}{r}\right) \text{d} t^2 + \left(1-\frac{R_{\text{S}}}{r}\right)^{-1} \text{d} r^2 + r^2 (\text{d} \theta^2 + \sin^2\theta \,\text{d} \phi^2).
\end{equation}
The Christoffel symbols are
\begin{align*}
\Gamma^t_{tr} &= \frac{R_{\text{S}}}{2r(r-R_{\text{S}})}, & \Gamma^r_{tt} &= \frac{R_{\text{S}}}{2r^3}(r-R_{\text{S}}), & \Gamma^r_{rr} &= \frac{-R_{\text{S}}}{2r(r-R_{\text{S}})}, \\
\Gamma^\theta_{r\theta} &= \frac{1}{r}, & \Gamma^r_{\theta\theta} &= -(r-R_{\text{S}}), & \Gamma^\phi_{r\phi} &= \frac{1}{r}, \\
\Gamma^r_{\phi\phi} &= -(r-R_{\text{S}}) \sin^2 \theta, & \Gamma^\theta_{\phi\phi} &= -\sin\theta \cos\theta, & \Gamma^\phi_{\theta\phi} &= \frac{\cos\theta}{\sin\theta}.
\end{align*}
The Killing equations are
\begin{align}
\tag{1a} &K_{t,t} - \Gamma^r_{tt} K_r = 0, \\
\tag{1b} &K_{t,r} + K_{r,t} - 2 \Gamma^t_{tr} K_t = 0, \\
\tag{1c} &K_{t,\theta} + K_{\theta,t} = 0, \\
\tag{1d} &K_{t,\phi} + K_{\phi,t} = 0, \\
\tag{1e} &K_{r,r} - \Gamma^r_{rr} K_r = 0, \\
\tag{1f} &K_{r,\theta} + K_{\theta,r} - 2 \Gamma^\theta_{r\theta} K_\theta = 0, \\
\tag{1g} &K_{r,\phi} + K_{\phi,r} - 2 \Gamma^\phi_{r\phi} K_\phi = 0, \\
\tag{1h} &K_{\theta,\theta} - \Gamma^r_{\theta\theta} K_r = 0, \\
\tag{1i} &K_{\theta,\phi} + K_{\phi,\theta} - 2 \Gamma^\phi_{\theta\phi} K_\phi = 0, \\
\tag{1j} &K_{\phi,\phi} - \Gamma^r_{\phi\phi} K_r - \Gamma^\theta_{\phi\phi} K_\theta = 0.
\end{align}
From ($1$e) we have
\begin{equation}\tag{2}
K_r = T(t,\theta,\phi) \left( \frac{r}{r-R_{\text{S}}} \right)^{1/2}.
\end{equation}
Differentiating ($1$b) with respect to $t$ and substituting ($1$a), ($1$e) and ($2$) into the result, we obtain
\begin{equation}\tag{3}
\left( \frac{\partial}{\partial r} \Gamma^r_{tt} + 3 \Gamma^r_{tt} \Gamma^r_{rr} \right) T(t,\theta,\phi) + \frac{\partial^2}{\partial t^2} T(t,\theta,\phi) = 0.
\end{equation}
Since
\begin{equation}
\frac{\partial}{\partial r} \Gamma^r_{tt} + 3 \Gamma^r_{tt} \Gamma^r_{rr} = \frac{R_{\text{S}}}{r^3}\left( -1 + \frac{3R_{\text{S}}}{4r} \right)
\end{equation}
is a function of $r$ only, and is not identically zero, ($3$) holds only when $T(t,\theta,\phi) \equiv 0$. Then we have $K_r \equiv 0$. So we can simplify the Killing equations to
\begin{align}
\tag{4a} &K_{t,t} = 0, \\
\tag{4b} &K_{t,r} = 2 \Gamma^t_{tr} K_t, \\
\tag{4c} &K_{t,\theta} + K_{\theta,t} = 0, \\
\tag{4d} &K_{t,\phi} + K_{\phi,t} = 0, \\
\tag{4e} &K_{\theta,r} = 2 \Gamma^\theta_{r\theta} K_\theta, \\
\tag{4f} &K_{\phi,r} = 2 \Gamma^\phi_{r\phi} K_\phi, \\
\tag{4g} &K_{\theta,\theta} = 0, \\
\tag{4h} &K_{\theta,\phi} + K_{\phi,\theta} = 2 \Gamma^\phi_{\theta\phi} K_\phi, \\
\tag{4i} &K_{\phi,\phi} = \Gamma^\theta_{\phi\phi} K_\theta.
\end{align}
From ($4$a), ($4$b), ($4$e), ($4$f) and ($4$g) we have
\begin{align}
K_t &= A(\theta,\phi) \left( 1 - \frac{R_{\text{S}}}{r} \right), \\
K_\theta &= B(t,\phi) \,r^2, \\
K_\phi &= C(t,\theta,\phi) \,r^2.
\end{align}
Substituting these results into ($4$c) and ($4$d), we obtain
\begin{align}
\frac{\partial}{\partial\theta} A(\theta,\phi) \left( 1 - \frac{R_{\text{S}}}{r} \right) + \frac{\partial}{\partial t} B(t,\phi) \,r^2 = 0, \\
\frac{\partial}{\partial\phi} A(\theta,\phi) \left( 1 - \frac{R_{\text{S}}}{r} \right) + \frac{\partial}{\partial t} C(t,\theta,\phi) \,r^2 = 0.
\end{align}
These equations hold only when $A$ is a constant and $B$ and $C$ are independent of $t$. So we have
\begin{align}
K_t &= A \left( 1 - \frac{R_{\text{S}}}{r} \right), \\
K_\theta &= B(\phi) \,r^2, \\
K_\phi &= C(\theta,\phi) \,r^2.
\end{align}
Substituting $K_\theta$ and $K_\phi$ into ($4$h) and ($4$i), we obtain
\begin{align}
\frac{\partial B}{\partial\phi} + \frac{\partial C}{\partial\theta} &= 2 \frac{\cos\theta}{\sin\theta} \,C, \\
\frac{\partial C}{\partial\phi} &= -\sin\theta \cos\theta \,B.
\end{align}
We can easily solve these PDEs to get
\begin{align}
B(\phi) &= -D \sin\phi + E \cos\phi, \\
C(\theta,\phi) &= -\sin\theta \cos\theta (D \cos\phi + E \sin\phi) + F \sin^2\theta,
\end{align}
where $D,E,F$ are constants. In summary, we have
\begin{align}
K_t &= A \left( 1 - \frac{R_{\text{S}}}{r} \right), \\
K_r &= 0, \\
K_\theta &= (-D \sin\phi + E \cos\phi) \, r^2, \\
K_\phi &= [-\sin\theta \cos\theta (D \cos\phi + E \sin\phi) + F \sin^2\theta] \, r^2.
\end{align}
The general solution of the Killing equations of the Schwarzschild metric will be
\begin{equation}
\begin{split}
K &= g^{\mu\nu} K_\mu \partial_\nu \\
&= g^{tt} K_t \frac{\partial}{\partial t} + g^{\theta\theta} K_\theta \frac{\partial}{\partial \theta} + g^{\phi\phi} K_\phi \frac{\partial}{\partial \phi} \\
&= - A \frac{\partial}{\partial t} +(-D \sin\phi + E \cos\phi) \frac{\partial}{\partial \theta} + [ - \cot\theta (D \cos\phi + E \sin\phi) + F ] \frac{\partial}{\partial \phi} \\
&= - A L_{(0)} + D L_{(1)} + E L_{(2)} + F L_{(3)},
\end{split}
\end{equation}
where
\begin{align}
L_{(0)} &= \frac{\partial}{\partial t}, \\
L_{(1)} &= -\sin\phi \frac{\partial}{\partial \theta} - \cot\theta \cos\phi \frac{\partial}{\partial \phi}, \\
L_{(2)} &= \cos\phi \frac{\partial}{\partial \theta} - \cot\theta \sin\phi \frac{\partial}{\partial \phi}, \\
L_{(3)} &= \frac{\partial}{\partial \phi}
\end{align}
form a basis of the Lie algebra of the Killing fields of Schwarzschild metric. $L_{(0)}$ is a timelike Killing vector field that is orthogonal to a foliation of spacelike hypersurfaces, representing a static spacetime. $L_{(1)}, L_{(2)}, L_{(3)}$ are Killing fields of a 2-sphere, representing a spacetime with a SO(3) symmetry.
