Is a spherically symmetric space-time isometric to a warped product?

Question

A spherically symmetric spacetime is a Lorentian 4-dimensional manifold $(M, g)$ whose isometry group contains a subgroup $G$ isomorphic to $\text{SO}(3)$ and whose orbits are 2-spheres. Here I am already confused. How can the orbits of $G$, which are submanifolds of $M$, be 2-spheres? Am I missing some general definition of 2-spheres that does not require being in Euclidean space? Next, how can we rigorously decompose the metric of this spacetime into one of the form $$ g = A(r, t) \text d r ^2 + B(r, t) \text d r \text d t + C(r, t) \text d t^2 + C(r, t) \text d \Omega ^2 \ ? $$ I am willing to take for granted that $\text{Iso}(M)$ is a Lie group and thus $G$ is diffeomorphic to $\text{SO}(3)$. In particular, the Lie algebra of $G$ is isomorphic to that of $\text{SO}(3)$ which is isomorphic to $(\Bbb R^3, \times)$. Because the Lie algebra of the isometry group consists of Killing fields, this means we have 3 Killing Fields $V_i$ s.t. $$ [V_i, V_j] = \epsilon_{ijk} V_k. $$ By Frobenius theorem, these vector fields generate foliation of $M$, and around every point of $M$ we can find a coordinate chart $(U, x^i)$ s.t. each leaf of our foliation corresponds to slices of constant $x^i$ with, say, $i = 0, 1$. Now I somehow need to find a coordinate transformation that ensures the vector fields $\partial_0$ and $\partial_1$ are orthogonal to every leaf of the foliation. To show that the inner products of the coordinate vector fields are independent of their location on each leaf I imagine I have to use the isometry conditions, but I do not know how the coordinate vector fields relate to the isometries themselves. Is there a canonical way to do this?