I am trying to prove the following statement:
A subset $K$ of an $m$-dimensional $M$ is an embedded submanifold of dimension $k$ if and only if around each $p\in K$, there exists a chart $(U,\phi)$ of $M$ such that \begin{align*} \left.\phi\right|_{U\cap K}:U\cap K \rightarrow \phi(U)\cap\left(\mathbb{R}^k\times \{0\}\right) \subset \mathbb{R}^m, \end{align*} which we call a submanifold chart of $K$.
An embedded submanifold of $M$ is defined as the image of an injective immersion $f:N \rightarrow M$ which is homeomorphic onto its image, and $f$ being an immersion means that the derivative $D_pf$ is injective at each point.
I have an attempt at proving the 'if' direction, but I'm very stuck with the other way. I'll put my attempt first, and then explain what I have tried going the other way.
Firstly, consider such a chart exists. We want to show that there exists a map from some manifold $N$, $f:N\rightarrow M$, such that $f(N)=K$ and $f$ is an injective immersion. Let $p\in K$. Then \begin{align*} \left.\phi\right|_{U\cap K}(p)=(\phi_1(p),...,\phi_m(p))\cap\left(\mathbb{R}^k\times \{0\}\right)=(\phi_1(p),...,\phi_k(p),0,...,0). \end{align*} Now we can compose this with the projector \begin{align*} \pi_K:\mathbb{R}^m&\rightarrow \mathbb{R}^k\\ (x_1,...,x_m)&\mapsto (x_1,...,x_k) \end{align*} to get \begin{align*} \pi_K \circ \left.\phi\right|_{U\cap K}(p)=(\phi_1(p),...,\phi_k(p)). \end{align*} However, note that this map is invertible. In particular, consider the inclusion map \begin{align*} i_K:\mathbb{R}^k&\rightarrow \mathbb{R}^k \times \{0_\mathbb{R}^{m-k}\}\\ (x_1,...,x_k)&\mapsto (x_1,...,x_k,0,...,0). \end{align*} Clearly $i_K\circ\pi_K=\text{id}_{ \mathbb{R}^k \times \{0_\mathbb{R}^{m-k}\}}$and $\pi_K \circ i_K= \text{id}_{\mathbb{R}^k}$.
We can use these maps to define a chart on $K$. We define \begin{align*} \phi_K^{-1}:=\left(\left.\phi\right|_{U\cap K}\right)^{-1}\circ i_K :\mathbb{R}^k&\rightarrow U\cap K\\ (x_1,...,x_k)&\mapsto\left(\left.\phi\right|_{U\cap K}\right)^{-1}((x_1,...,x_k,0,...,0)), \end{align*} where we have used the fact that $\phi$ is an invertible map since it forms a chart of $M$. $\phi_K^{-1}$ is invertible, and we define it as \begin{align*} \phi_K:=\pi_K \circ \left.\phi\right|_{U\cap K}. \end{align*} But this is a chart of $U \cap K$. In particular, it is the composition of differentiable functions. Hence, we define the map \begin{align*} f=\left(\left.\phi\right|_{U\cap K}\right)^{-1}\circ i_K \circ \phi_K: K \supset U\cap K \rightarrow U\cap K \subset M, \end{align*} which is a differentiable map from $K\rightarrow M$. Although it is only defined for some open set $U$, we can do the same construction over the entire manifold (and any overlap must be compatible since it is inherited from the overlap of $\phi$). We see that $N=K$ as a manifold, as we were able to equip the subset $K$ with a manifold structure. Then $f$ maps to $K$ as a subset of $M$. To show that $K$ is an embedded submanifold, we need to show that $f$ is an injective immersion. Let $p,p'\in K$. Then \begin{align*} &&f(p)&=f(p')\\ &\iff&\left(\left.\phi\right|_{U\cap K}\right)^{-1}\circ i_K \circ\pi_K \circ \left.\phi\right|_{U\cap K}(p)&=\left(\left.\phi\right|_{U\cap K}\right)^{-1}\circ i_K \circ\pi_K \circ \left.\phi\right|_{U\cap K}(p')\\ &\iff&\left(\left.\phi\right|_{U\cap K}\right)^{-1}\circ \text{id}_{\mathbb{R}^m} \circ \left.\phi\right|_{U\cap K}(p)&=\left(\left.\phi\right|_{U\cap K}\right)^{-1}\circ \text{id}_{\mathbb{R}^m} \circ \left.\phi\right|_{U\cap K}(p')\\ &\iff&\left(\left.\phi\right|_{U\cap K}\right)^{-1}((\phi_1(p),...,\phi_k(p),0,...,0))&=\left(\left.\phi\right|_{U\cap K}\right)^{-1}((\phi_1(p'),...,\phi_k(p'),0,...,0))\\ &\iff&p&=p', \end{align*} since $\phi$ is bijective. So $f$ is injective. To show $f$ is an immersion, choose the charts $(U\cap K,\left.\phi\right|_{U\cap K})$ for $M$ and $(U\cap K,\phi_K)$ for $K$. In these charts, \begin{align*} &D_{f(p)}\left.\phi\right|_{U\cap K} \circ D_pf \circ D_{\phi_K(p)}\phi_K^{-1}([\phi_K \circ \gamma])\\ =&D_{f(p)}\left.\phi\right|_{U\cap K} \circ D_pf\left([\phi_K^{-1}\circ\phi_K \circ \gamma] \right)\\ =&D_{f(p)}\left.\phi\right|_{U\cap K}\left([f\circ\phi_K^{-1}\circ\phi_K \circ \gamma]\right)\\ =&[\left.\phi\right|_{U\cap K}\circ f\circ\phi_K^{-1}\circ\phi_K \circ \gamma]\\ =&D_{\phi_K(p)}\left(\left.\phi\right|_{U\cap K}\circ f\circ\phi_K^{-1}\right)([\phi_K \circ \gamma])\\ =&D_{\phi_K(p)}i_K([\phi_K \circ \gamma])\\ =&\left.\frac{d}{dt}\right|_{t=0}\left(i_k\circ \phi_K \circ \gamma\right)\\ =&\left.\frac{d}{dt}\right|_{t=0}((\phi_1\circ \gamma),...,(\phi_k\circ\gamma),0,...,0)\\ =&\left(\dot{(\phi_1\circ\gamma)}(0),...,\dot{(\phi_1\circ\gamma_k)(0)},0,...,0\right) \end{align*} and \begin{align*} [\phi_K \circ \gamma]&=\left.\frac{d}{dt}\right|_{t=0}\left(\phi_K\circ \gamma\right)\\ &=\left.\frac{d}{dt}\right|_{t=0}((\phi_1\circ \gamma),...,(\phi_k\circ\gamma))\\ &=\left(\dot{(\phi_1\circ\gamma)}(0),...,\dot{(\phi_1\circ\gamma_k)(0)}\right), \end{align*} meaning \begin{align*} D_{\phi_K(p)}i_K= \begin{pmatrix} 1&0&...&0&0&...&0\\ 0&1&...&0&0&...&0\\ \vdots&\vdots&\ddots&\vdots&\vdots&...&\vdots\\ 0&0&...&1&0&...&0\\ \end{pmatrix}, \end{align*} which acts on row vectors with right multiplication. In particular, let \begin{align*} (a_1,\dots,a_k),(b_1,\dots,b_k)\in T_{\phi_K(p)}\mathbb{R}^k. \end{align*} Then \begin{align*} &&D_{\phi_K(p)}i_K((a_1,\dots,a_k))&=D_{\phi_K(p)}i_K((b_1,\dots,b_k))\\ &\iff&(a_1,\dots,a_k,0,\dots,0)&=(b_1,\dots,b_k,0,\dots,0)\\ &\iff&(a_1,\dots,a_k)&=(b_1,\dots,b_k). \end{align*} So we have proved the existence of a flattener of a subset means that the subset is also an embedded submanifold.
To go the other way, we assume that $K$ is an embedded submanifold of $M$. This means $K$ is the image of an injective immersion $f:N\rightarrow M$, for some manifold $N$. We want to show that there exists a chart $(U,\phi)$ around each $p\in K$ such that $\left.\phi\right|_{U\cap K}$ is a flattener. This is where I am getting stuck. There are two theorems that I think may be useful:
Sard's Theorem: For any differentiable map $f:M\rightarrow N$ between smooth manifolds, the set of regular values is dense in $N$.
Regular Point Theorem: Let $p$ be a regular point of the map $f$. Then there exists charts $(U, \phi)$ of $M$ around $p$ and $(V,\psi)$ of $N$ around $f(p)$ with \begin{align*} &\phi(p)=0\\ &\psi(f(p))=0\\ &f(U)\subset V \end{align*} such that \begin{align*} \psi \circ f \circ \phi^{-1}(x_1,\dots,x_{n+k})=(x_1,\dots,x_n), \end{align*} where $\text{dim}M=n+k$ and $\text{dim}N=n$.
(Note that I want to avoid using the regular value theorem if that is a way to prove this) The only idea I had was that maybe $f$ is somehow a submersion, and then I could use the charts in the regular point theorem, and maybe I need Sard's theorem to show that. But I couldn't get anywhere, so maybe that's wrong.