Let $M$ be a smooth manifold, and $G$ be a compact Lie group acting on $M$ smoothly.
We assume that isotropic group of any point $x\in M$ is of dimension zero.
Q How to show that $M$ admits a smooth foliation with dimension $\dim (G)$, whose leaf is the orbit $Gx$ for any $x\in M$.
PS: I saw this on some talk, but I did figure out where to use the condition isotropic group is of dimension zero. Any reference or hint is welcome.
I'll be using the definition of foliation that appears in Lee's book (page 501, it should be equivalent to the definition that you are using). Essentially, a foliation of dimension $k$ is a partition of $M$ into connected $k$-dimensional immersed submanifolds which locally look like the union of hyperplanes. Also, I'll be assuming that $G$ is connected and $k$-dimensional.
The proof that I propose uses quite a lot of machinery from distributions, so if you're not familiar with those feel free to ask!
First, there is a correspondence between foliations and (involutive) distributions: if $\mathcal{F}=\{A_{p}\colon p\in M\}$ is a $k$-dimsntional foliation of $M$ (and $A_{p}$ denotes the leaf which contains $p$), then the subset $\Delta=\bigcup_{p\in M}T_{p}(A_{p})$ of $T(M)$ must be an involutive rank $k$ distribution. Conversely, if $\Delta=\bigcup_{p\in M}\Delta_{p}$ is an involutive distribution, for every $p\in M$ there is a unique maximal connected integral manifold of $\Delta$ passing through $M$, and the collection of all integral manifolds is a $k$-dimensional foliation of $M$. These results are due to the Frobenius Theorem, which is wonderfully explained in "Foundations of Differentiable Manifolds and Lie groups", by Warner (pages 41-49). Also, it's important to know that integral manifolds of an involutive distribution are weakly embedded (Warner, Theorem 1.62).
So, suppose $\{G\cdot p \colon p\in M\}$ is a foliation. Then, $\Delta=\bigcup_{p\in M}T_{p}(G\cdot p)$ must be an involutive distribution. For this to be possible, we need to guarantee that $\dim T_{p}(G\cdot p)$ is independent of $p$ (that is, all orbits must have the same dimension, which is equivalent to saying that all isotropy subgroups must have the same dimension).
Let's go now to the question at hand: we're assuming that $G_{p}$ is zero-dimensional for all $p$. This implies that $\dim G\cdot p = \dim G$ for all $p\in M$, so $\Delta=\bigcup_{p\in M}T_{p}(G\cdot p)$ is a union of $k$-subspaces. We need it to be a distribution. For this, take any basis $\{X_{1},...,X_{k}\}$ of the Lie algebra $\mathfrak{g}$ of $G$ and define
$$ X_{i}^{*}(p)=\dfrac{d}{dt}\Big|_{t=0}\operatorname{Exp}(tX_{i})\cdot p, \quad p\in M. $$
The vector fields $X_{i}^{*}$ are well defined and smooth in $M$, and they form a basis for $T_{p}(G\cdot p)$ at every point $p\in M$ (notice that this is still true locally if we suppose that the orbits have constant dimension. We'd have that the fields span those subspaces). From this, we get that $\Delta$ is a distribution. Also, it is involutive because the orbits are integral manifolds. The Frobenius Theorem (global version) now implies that the collection $\mathcal{F}=\{A_{p}\colon p\in M\}$ of maximal integral manifolds of $\Delta$ is a $k$-dimensional foliation of $M$. In general, since $G\cdot p$ is a connected integral manifold through $p$, $G\cdot p\subseteq A_{p}$ (because all conected integral manifolds are contained in the maximal integral manifold, c.f. Warner Theorem 1.64), so if we can guarantee that $G\cdot p=A_{p}$, we are done.
By hypothesis, $G$ is compact, so the action is proper. In particular, the orbits are closed in $M$ (c.f. Lee, Proposition 21.7). Now, let $p\in M$. We know that $G\cdot p\subseteq A_{p}$ and it is closed in $M$, hence $G\cdot p$ is closed in $A_{p}$. This is because the inclusion $A_{p}\hookrightarrow M$ is continuous, but not neccessarily an embedding. Now, consider the inclusion $G\cdot p \hookrightarrow A_{p}$. Since $A_{p}$ is weakly embedded, this inclusion is a smooth immersion. Also, $G\cdot p$ and $A_{p}$ have the same dimension, so the inclusion is a submersion and $G\cdot p$ is open in $A_{p}$. Now, $A_{p}$ is connected, se we get that $G\cdot p=A_{p}$, and we conclude.
Hope this helps!
