Submanifolds and adapted atlas Let $M$ be a smooth manifold of dimension $n$. 
My notes say

Theorem: A subset $S$ of $M$  could be given a structure of smooth manifold of dimension $k$ such that $S$ is an embedded submanifold of $M$ (i.e. the inclusione map $\iota:S\hookrightarrow M$ is a smooth embedding) if and only if for each point $p$ in $M$ there exists a smooth chart $(U,\phi)$ for $M$ centered at $p$ and $k$-adapted to $S$ (i.e. $U\cap S=\emptyset$ or $\phi(U\cap S)=\{x\in \phi(U):x^{k+1}=\dots=x^n=0\}$)(Or equivalently: there exists a smooth atlas for $M$ which is $k$-adapted to $S$, meaning that for each smooth chart $(U,\phi)$ in that altas, I have $(U,\phi)$ is $k$ adapted to $S$). 

Now, in the proof we only show that if I take a point $p$ in $S$ (not, in general, in $M$ as stated above!) then exists a smooth chart $(U_p,\phi_p)$ for $M$ centered at $p$ and $k$-adapted to $S$. But now if I consider $\{(U_p,\phi_p)\}_{p\in S}$ I could not have an atlas for $M$. (I only know that $S\subseteq\bigcup_{p\in S}U_p$ but not that $M=\bigcup_{p\in S}U_p$).
So, the statement of the above theorem is not very correct or am I missing something? How could I complete (if I can!) the set $\{ (U_p,\phi_p)\}_{p\in S}$ to obtain a smooth atlas for $M$ such that each chart is $k$-adapted to $S$ ?
Or should I modify the statement in 


Theorem: A subset $S$ of $M$  could be given a structure of smooth manifold of dimension $k$ such that $S$ is an embedded submanifold of $M$ if and only if for each point $p$ in $S$ there exists a smooth chart $(U,\phi)$ for $M$ centered at $p$ and $k$-adapted to $S$ ?


 A: The correct statement is that $S \subset M$ is a closed, embedded submanifold if and only if $M$ has an atlas of adapted charts. If I were being careful / if $S$ wasn't embedded a better way to phrase this would be to start by writing $f: S \to M$, which I state is a proper injective immersion.
The essential input into constructing such an atlas is the following.
1) Given any $x \in S$, choose a small chart on $S$ with domain $V$; then there is an open subset $U \subset M$ containing $x$ so that $U \cap S \subset V$. (That is, "far away points of $S$" from the perspective of the manifold topology on $S$ do not accumulate towards some fixed point $f(x)$ in the image.) 
2) Given any $x \in M \setminus S$, there should be a neighborhood $U \subset M$ containing $X$ so that $U \cap X = \varnothing$. 
The first is necessary so that $f: S \to f(S) \subset M$ is a homeomorphism, where we consider $f(S)$ with the subspace topology (which is guaranteed by the existence of adapted charts near points of $S$). The second is necessary to get adapted charts away from $S$. 

(2) is equivalent to saying that $S$ is a closed subset of $M$. 
(1) is harder: it says that if there is a sequence $x_n \in S$ and $x \in S$ so that $f(x_n) \to f(x)$, then in fact we must have $x_n$ near $x$; that is, there is a subsequence of $x_n$ which converges to $x$. This is saying that the embedding map is "proper": the inverse image of compact sets, like $\{f(x_1), f(x_2), \cdots, f(x)\}$, is compact. 
But if $f$ is a proper embedding (= proper injective immersion = injective immersion with closed image), then one may indeed find adapted charts. 
The definitions already almost tell us the proof. If $p \in S$, we know that we may find some chart $V \subset S$ of $p$ and some chart $U \subset M$ of $f(p)$ so that on these charts, the map $f: V \cap U \to U$ is given by the inclusion of $\Bbb R^k$ into $\Bbb R^n$, by the implicit function theorem. What we don't know is that $U \cap S = V \cap S$, so that $S$ "never otherwise appears in $U$". This is where your proof needs the properness assumption. This gives an adapted chart to any point $p \in S$.
If $p \not \in S$, then by assumption we may choose a chart with domain $U$ having trivial intersection with $S$. If you already had a chart with domain $U'$, then the new  chart is the same map on the subdomain $U = U' \cap (M \setminus S)$, which is again open.
