Quoting Tu's "An introduction to manifolds":
Definition 9.1. A subset $S$ of a manifold $N$ of dimension $n$ is a regular submanifold of dimension $k$ if for every $p \in S$ there is a coordinate neighborhood $(U,f) = (U,x_1,\dots,x_n)$ of $p$ in the maximal atlas of $N$ such that $U \cap S$ is defined by the vanishing of $n−k$ of the coordinate functions. By renumbering the coordinates, we may assume that these $n−k$ coordinate functions are $x_{k+1},\dots,x_n$. We call such a chart $(U,f)$ in $N$ an adapted chart relative to $S$. On $U \cap S$, $f = (x_1, . . . ,x_k,0, . . . ,0)$. More concretely,
$$f(U\cap S) = f(U) \cap (\mathbb{R}^k \times \{0_{n-k}\})$$ Let $f_S : U\cap S→\mathbb{R}^k$ be the restriction of the first $k$ components of $f$ to $U \cap S$, that is, $f_S = (x_1, . . . ,x_k)$. Note that $(U\cap S,f_S)$ is a chart for $S$ in the subspace topology.
My notes then say that if we put the subspace topology on $S$ and the atlas $\{(U\cap S, f_S)\}$, then $S$ becomes a smooth manifold with as maximal atlas the unique atlas containing $\{(U\cap S, f_S)\}$.
I'm trying to prove this (Tu's book contains a proof, but this seems easier).
Here is my idea:
Clearly $U \cap S$ is open in $S$, because $U$ is open in $M$.
Define $\operatorname{pr}: f(U \cap S) \to pr(f(U \cap S)): (x_1, \dots, x_n) \mapsto (x_1, \dots, x_k)$, thus the projection that forgets the last $n-k$ coordinates.
Note that on the given domain and codomain, $pr$ is bijective with
$$\operatorname{pr}^{-1}(x_1, \dots, x_k) = (x_1, \dots, x_k, 0, \dots, 0)$$
Then, observe that $f_S = \operatorname{pr} \circ f$. Clearly, $pr$ is a $C^\infty$ diffeomorphism on the given domain and codomain, and thus $f_S$ is a homeomorphism as composition of homeomorphisms.
I also have a proof that $f_S(U \cap S)$ is an open subset of $\mathbb{R}^k$, but to save space I won't include it here.
Thus, it remains to prove that any two charts $f_S: U \cap S\to \mathbb{R}^k, g_S: U \cap S \to \mathbb{R}^k$ are compatible. This is clear, because on $f_S(U \cap V \cap S)$ we have
$$g_S \circ f_S^{-1} = (\operatorname{pr}^g \circ g) \circ (\operatorname{pr}^f \circ f)^{-1} = \operatorname{pr}^f \circ (g \circ f^{-1}) \circ \operatorname{pr}^f$$ which is $C^\infty$, as composition of $C^\infty$ maps.
Is the above correct? Thanks in advance for any help.