Introduction to Smooth Manifolds by John M. Lee:
Lemma 6.13
Suppose $M\subset \mathbb{R}^N$ is a smooth $n$-dimensional submanifold with or without boundary. For any $v\in \mathbb{R}^N \setminus \mathbb{R}^{N-1}$, let $\pi_v: \mathbb{R}^N \to \mathbb{R}^{N-1}$ be the projection with kernel $\mathbb{R}v$ (where we identify $\mathbb{R}^{N-1}$ with the subspace of $\mathbb{R}^N$ consisting of points with last coordinate zero). If $N>2n+1$, then there is a dense set of vectors $v\in \mathbb{R}^N \setminus \mathbb{R}^{N-1}$ for which $\pi_v\vert_M$ is an injective immersion of $M$ into $\mathbb{R}^{N-1}$.
Lemma 6.14
Let $M$ be a smooth $n$-manifold with or without boundary. If $M$ admits a smooth embedding into $\mathbb{R}^N$ for some $N$, then it admits a proper smooth embedding into $\mathbb{R}^{2n+1}$.
When proving Lemma 6.14, the first initial idea we get is that we should just compose our embedding $f$ with $\pi_v$ and get $\pi_v\circ f$ an injective immersion into $\mathbb{R}^{N-1}$, which might be an embedding, and then repeat the process. I could prove that it is in fact an embedding, but it is not necessarily proper. Professors Lee's proof is a lot longer and uses a very non trivial map, and of course we get a proper map (as promised). The length of the proof makes me doubt that my proof is correct. If it is correct, then it would mean that only the properness makes the whole thing so difficult. Please check if my thoughts work (for the non-proper case):
My proof:
By iterating, it is enough to show that $\pi_v\vert_M \circ f$ is a smooth embedding for some $v\in \mathbb{R}^N \setminus \mathbb{R}^{N-1}$. By composition, $\pi_v\vert_M \circ f$ is smooth, an immersion, and injective.
The only thing we have to check is that it is a topological embedding.
We know that $\pi_v\vert_M \circ f$ is continuous into $\mathbb{R}^{N-1}$, so it is also continuous onto $(\pi_v\vert_M \circ f)(M)$.
The very last thing to show, is that $\pi_v\vert_M \circ f$ maps open subsets of $M$ onto open subsets of $(\pi_v\vert_M \circ f)(M)$ in the subspace topology.
Let $O\in \tau_M$ open in $M$. Since $f$ is an embedding, $$f(O)=U \cap f(M),$$ where $U$ open in $\mathbb{R}^N$. Then $$\pi_v\vert_M f(O) = \pi_v(U) \cap \pi_v\vert_M f (M)$$ is open in the subspace topology of $(\pi_v\vert_M \circ f)(M)$, since $\pi_v (U)$ open in $\mathbb{R}^{N-1}$