Integral Manifolds of the Symplectic Foliation Let $(M,\pi)$ be a Poisson manifold. Denote by $\pi^*:T^*M \rightarrow TM$, the induced bundle map. In the simple case, where the Poisson bivector is of constant rank, we obtain a smooth regular foliation of M, i.e. $Im\pi^*$, whose leaves are actually symplectic submanifolds. In the case that $\pi$ is not of constant rank, we obtain a (singular) symplectic foliation of the underlying manifold. I am trying to figure out the details of this foliation.
The construction of the leaves for the singular foliation is the following:

*

*First one observes that the distribution $Im(\pi^{*})$ has integral manifolds. Let $x\in M$ and $\big(U,(p_i,q^i,x^s)\big)$ be the Darboux-Weinstein coordinates centered at $x$. Then, the submanifold $S=\{x^s=0\}$ is an integral submanifold containing x.


*The second step is to show that the integral manifolds of $Im(\pi^*)$ are actually weakly embedded submanifolds and that the connected components of the intersection of two integral manifolds are also integral manifolds.


*Lastly, for a point $x_0 \in M$ we take the union of all integral manifolds containing it, and use a gluing lemma for weakly embedded submanifolds to show that this union is the maximal integral manifold containing $x_0$, i.e the symplectic leaf of the foliation that contains $x_0$.

In order to prove the second step, we use the following lemma:
Lemma: Let N be an integral manifold, $x\in N$ and $\big(U,(p_i,q^i,x^s)\big)$ the Darboux-Weinstein coordinates at $x$. Then the following hold:
i)The connected components of $N \cap U$ are contained in the slices $\{x^s=constant\}$.
ii) If $N'$ is another integral manifolds containing $x$ then the connected component of $N\cap N'$ containg $x$ is an integral manifold.
I am stack at the two arguments that prove the Lemma:
For i) it is  claimed that it suffices to show that $N\cap \{x^s=constant\}$ is both open and closed in $N\cap U$. I do not see why this is sufficient. Then, it is claimed that ii) follows from i) since the slices $\{x^s=const.\}$ and integral manifolds have the same dimension. My intuition on this tells me that we are fine, since the tangent spaces at the intersection overlap nicely but what is the smooth manifold structure on $N \cap N'$ ?
Any help to clear things out is deeply appreciated!
 A: To i): a set is connected if and only if the only open and closed subsets are the empty set and the set itself. Hence if $N\cap \{x^s=c\}$ is open and closed in $N\cap U$ then any connected subset of $N\cap U$ is contained in some $N\cap \{x^s=c\}$, in particular the maximal connected sets are, which are the connected components.
To ii): we know from i) that the connected component of $N\cap N'\cap U$ which contains $x$ is contained in $\{x^s=0\}$. Denote by $y$ the coordinates of in $N$ around $x\in N$ and similar let $z$ denote the coordinates around $x\in N'$. Integral manifolds are in particular immersed submanifolds. Denote the immersion of $N \hookrightarrow M$ by $\phi_{N}$ and of $N' \hookrightarrow M$ by $\phi_{N'}$. We can assume that by the local immersion theorem that $\phi_{N}(y)=(y_i,y^{i+n},0)_{1\le i \le n}$ where $n= \frac{1}{2} \mathrm{dim}(\mathrm{Im}(\pi^{\sharp}_x))$. Then $\phi_{N'}$ is locally around $x\in N'$ given by
\begin{align*}\phi(z)= (p_i(z) ,q^i(z),0)
\end{align*}
Hence one obtains a function from a neighorhood of $x\in N'$ to a neighborhood of $x\in N$ by
\begin{align*} z\mapsto (p_i(z) ,q^i(z))
\end{align*}
Running the same argument the other way around one can show that this is a local diffeomorphism. Hence the manifold structure on $N\cap N'$ can be chosen to be either one of the restrictions of the manifold structures.
