Adapted charts for smooth manifold Let $M$ be a smooth manifold of dimension $n$ and let $S$ be an embedded submanifold of $M$, i.e. a subset of $M$ which is given a structure of smooth manifold of dimension $k\leq n$ such that the inclusion map $\iota:S \to M$ is a smooth embedding.
By the constant rank theorem, if $p$ is a point of $S$, then there exists smooth charts $(W,\chi)$ for $S$ centered at $p$ and $(V,\psi)$ for $M$ centered at $p$ with $W \subseteq V$ such that the map $\psi\circ\chi^{-1}:\chi(W)\to\psi (V)$ has the expression $(x^1,\dots,x^k)\mapsto (x^1,\dots,x^k,0,\dots,0) $.
In particular the map $\psi\circ\chi^{-1}:\chi(W)\to\psi (W)$ is a bijection and thus $$\psi(W)=\chi(W)\times\{ \underline{0} \}$$
Since $W$ is open in $S$, we have $W=S\cap A$ with $A$ open subset of $M$.
Let be $U=V \cap A$. Then $U$ is open in $V$ and thus $\psi(U)$ is open in $\mathbb{R}^n$. We also have $U\cap S=W$. So we have $$\psi(U\cap S)=\chi(W)\times\{ \underline{0} \}\quad [1]$$

I want to say that $\psi(U\cap S)=\{ x\in\psi(U) : x^{k+1}=\dots=x^n=0\}$.
My notes say that I can suppose $\psi(U)=B^n_\varepsilon(0)$ and $\chi(W)=B^k_\varepsilon(0)$ be open balls in $\mathbb{R}^n$ and $\mathbb{R}^k$ respectively with the same radius but I can't see why.

I know that I can WLOG assume $\chi(W)=B^k_\varepsilon(0)$ for a certain $\varepsilon>0$ but then $U$ is determinated by $W$! (See the definition of $U$) So I can't modify $U$ and then still pretend that $[1]$ holds.

Can you tell me why the conclusion of my notes is true (if it is true!) or provide another strategy to conclude the argument?

 A: 
Let us understand what the claim is doing here.

Definition: Let $S \subseteq M$, $S$ satisfies the $k$-slice condition if each point of $S$ lies in smooth chart $(U, \varphi)$ for $M$ such that $S \cap U$ is a single $k$-slice in $U$, i.e. for a constant $c$, in coordinates,
  $$ S = \{ x \in U \, : \, x^{i}=c^i, \, k+1 \le i \le n\}$$ 


So what we want to show is thatany embedded manifold satisfies the condition. 
Your set equality is by definition: but let me explain the steps. 

Making the radius smaller is used in first changing your $W$ and $V$.   Choose open balls $B' \subseteq V$, $B \subseteq W$, such that the image of $B$ is precisely a $k$-sliceof $B'$. This is possible by making the radius of the balls sufficiently small. We may take these as our $W$ and $V$. 

One might think we are done here. That $V$ is our desired $k$-slice. i.e. $V \cap S=W$. Nonetheless, we have:
$$ W \subseteq V \cap S$$ 

The remedy is to introduce your open set $A$.  

As you wrote, $W = S \cap A$ for an open set $A$ in $M$. Then $U= A \cap V$ is our desired $k$-slice. This is a set chase.  One direction is already proven. Now $$ U \cap S = (A \cap V) \cap  S = (A \cap S) \cap V = V \cap W = W$$
In words, this means, 

$U \cap S$ is precisely the $k$-slice of $U$


At this point it is clear, in coordinates: if $(x,0) \in U \cap S$, then it lies both in $S$ and is in the $k$-slice of $U$. But above shows this implies it must lie in $W$. Hence $U \cap S \subseteq W$.  The converse is also clear, we know $W$ is represented as the $k$-slice in $U \cap S$. Thus your equality. 

In fact more is true. If a subset satisfies the condition, then it can be given a (unique) manifold structure such that it is an embedded $k$-manifold.
