How to prove a compact $n$ - dimensional manifold cannot be embbeded into $\mathbb{R}^n$ 
Let $M$ be an $n$ - dimensional $C^1 $compact manifold, i.e, for any cover of open sets in the topology there exists a finite cover. Prove that $M$ cannot be embedded into $\mathbb{R}^n$.

I tried to assume by contradiction that there exists embedding, meaning there exists a $C^1$ diffeomorphism $ M\overset{\varphi}{\longrightarrow}\varphi\left(M\right)\subseteq\mathbb{R}^{n} $, and I assume one can easily prove that the image $\varphi(M)$ is also compact in $\mathbb{R}^n$ since we can find a finite cover for any open cover. But I do not have an idea for a contradiction.
Any help would be appreciated. Thanks in advance.
 A: Following up on my comment that the only $n$-dimensional (embedded) submanifolds of $\Bbb{R}^n$ are non-empty open sets. This is just a matter of unwinding definitions.
For the 'non-trivial' direction, consider any point $x\in X$. By definition, there exist open sets $U,V\subset \Bbb{R}^n$  and a $C^1$ diffeomorphism $\Phi:U\to V$, such that $x\in U$ and $\Phi(U\cap X)=V\cap (\Bbb{R}^{\dim X}\times\{0_{\Bbb{R}^{n-\dim X}}\})$ (i.e that you can locally flatten $X$ to look like a piece of $\Bbb{R}^{\dim X}$). Now, since $n=\dim X$, this reduces to $\Phi(U\cap X)=V$. But recall that $V=\Phi(U)$ since $\Phi$ is a diffeomorphism (hence bijective), therefore, we have $U\cap X=U$, and hence $U\subset X$. Since the point $x$ was arbitrary, we have shown that $X$ is a union of open subsets of $\Bbb{R}^n$, and hence $X$ is open in $\Bbb{R}^n$.
Finally, a non-empty open set in $\Bbb{R}^n$ is not compact, because compact implies closed, so if a set is closed and open, then by connectedness of $\Bbb{R}^n$, the set is either $\emptyset$ (which we excluded) or $\Bbb{R}^n$ (which is not compact for $n\geq 1$).
A: Any point on the boundary (wrt to the surrounding topology) can't have a local coordinate map by Brouwer's invariance of domain theorem https://en.m.wikipedia.org/wiki/Invariance_of_domain
Feels like there should be a simpler solution though.
EDIT: Ah, you're talking $C^1$, in that case proving invariance of domain is easy
