Here's what I'm trying to prove;
Let $D \subseteq \mathbb{R}^d$ be a non-empty open set. Let $d < n$ and $k \in \mathbb{N}_{\infty}$. Let $\phi: D \to \mathbb{R}^n$ be a $C^k$ embedding. Then, $\phi(D)$ is a $C^k$ submanifold in $\mathbb{R}^n$ of dimension $d$.
The corollary is from Duistermaat and Kolk's Multidimensional Analysis I so I am using definitions and so on from that book.
Attempted Proof:
Observe that $\phi: D \to \phi(D)$ is a homeomorphism and that $D\phi(y)$ is injective for every $y \in D$. Let $x \in \phi(D)$. Then, there exists a unique $y \in D$ such that $\phi(y) = x$. By the Immersion Theorem, there exists an open set around $y$, call it $U_y$, such that $\phi(U_y)$ is a $C^k$ submanifold in $\mathbb{R}^n$ of dimension $d$.
Since $x \in \phi(U_y)$, there exists an open set $V_x$ around $x$ such that $\phi(U_y) \cap V_x$ is the graph of some $C^k$ function $f: \mathbb{R}^d \supset\to \mathbb{R}^{n-d}$ defined on an open set. Since $\phi$ is a homeomorphism, $\phi(U_y)$ is open and so, $\phi(U_y) \cap V_x$ is open around $x$. Then, observe that $\phi(D) \cap (\phi(U_y) \cap V_x) = \phi(U_y) \cap V_x$ is equal to the graph of some $C^k$ function (the same one as above). This proves that $\phi(D)$ is a $C^k$ submanifold in $\mathbb{R}^n$ of dimension $d$ at $x$. Since $x$ was arbitrary, we have our result. $\Box$
Does the argument work? If it doesn't, then why? How can I fix it?