Inverse images of coadjoint orbits are coisotropic submanifolds?

Problem Let $$(M, \omega)$$ be a symplectic manifold. Let $$G$$ be a connected, compact Lie group acting on $$M$$. Let $$J: M \rightarrow \mathfrak{g}^{*}$$ be the moment map. Let $$\eta$$ be a regular value of $$J$$ and let $$\mathcal{O}$$ be the orbit of $$\eta$$ under the coadjoint action, i.e. $$\mathcal{O} = \left\{Ad_{g^{-1}}^{*} \eta \mid g \in G \right\}$$.

Prove that $$i: J^{-1} (\mathcal{O}) \rightarrow M$$ is a coisotropic submanifold.

Attempt: Let $$p \in J^{-1} (\mathcal{O})$$. Then by definition, I have to show that $$T_p (J^{-1} (\mathcal{O}))^{\omega} \subset T_p (J^{-1} (\mathcal{O}))$$.

I know the following, that $$T_q (J^{-1} (\eta))^{\omega} = T_q (G \cdot q)$$ where $$G \cdot q = \left\{ \Phi(g,q) \mid g \in G \right\}$$ is the orbit and $$q \in J^{-1} (\eta)$$. Also, by standard differential geometry, since $$\eta$$ is a regular value, we have $$T_q J^{-1} (\eta) = \text{ker} (T_q J)$$.

Also, there is a result that $$T_q (G_{\eta} \cdot q) = T_q (G \cdot q) \cap T_q(J^{-1} (\eta)).$$ It does not follow that $$J^{-1} (\mathcal{O})$$ is a coistropic submanifold of $$M$$, if I would prove that for every $$\zeta \in \mathcal{O}$$, the inverse image $$J^{-1} (\zeta)$$ is a coisotropic submanifold (which might be false, not sure)?

So how do I figure out what $$T_p (J^{-1} (\mathcal{O}))^{\omega}$$ is?

1) You should first argue why $$J^{-1}(\mathcal{O})\subset M$$ is a submanifold in the first place. To do so, it is enough to remark that $$J:M\rightarrow\mathfrak{g}^{*}$$ is transverse to $$\mathcal{O}\subset\mathfrak{g}^{*}$$, i.e. for all $$q\in J^{-1}(\mathcal{O})$$ we have $$d_q J(T_q M)+T_{J(q)}\mathcal{O}=\mathfrak{g}^{*}.$$ This is the case because $$J(q)\in\mathcal{O}$$ is also a regular value, so that $$d_q J(T_q M)=\mathfrak{g}^{*}$$.
2) Now let $$p\in J^{-1}(\mathcal{O})$$ and assume that $$J(p)=\zeta\in\mathcal{O}$$, i.e. $$p\in J^{-1}(\zeta)$$. Since $$T_{p}J^{-1}(\zeta)\subset T_{p}J^{-1}(\mathcal{O})$$, we have $$(T_{p}J^{-1}(\mathcal{O}))^{\omega}\subset (T_{p}J^{-1}(\zeta))^{\omega}=T_{p}(G\cdot p).$$ So in order to conclude, it is enough to show that $$G\cdot p\subset J^{-1}(\mathcal{O})$$. This inclusion holds because $$J$$ is equivariant: $$J(g\cdot p)=Ad^{*}_{g}(J(p))\subset Ad^{*}_{g}(\mathcal{O})=\mathcal{O}.$$