Let $M$ be a compact submanifold of dimension $n$ in $\mathbb{R}^{n+1}$ show that exists a normal unitary smooth vector field in $M$ 
Let $M$ be a compact submanifold of dimension $n$ in $\mathbb{R}^{n+1}$ show that exists a normal unitary smooth vector field in $M$

First i think to these is equivalent to show that $M$ is orientable, then how $\dim(M)=n$ and $M$ is compact then i try to work with the Jordan - Brouwer separation theorem because these gives me a open set $A$ such that $\overline{A}$ is a compact submanifold and such that $\partial (\overline{A})=M$ but i not sure to these is correct.
Any hint or help i will be very grateful
 A: It seems like people suggested in the comments that orientability should be a hypothesis. I don’t think the problem assumes orientability. In fact, it seems like the purpose is to prove that a codimension $1$ submanifold of $\mathbb{R}^{n+1}$ must be orientable.
Assume for a contradiction $M$ is not orientable. Then there exists a loop $\gamma$ on $M$ such that it is not possible to choose consistently oriented bases on all coordinate patches simultaneously that intersect $\gamma$. By picking an arbitrary orientation on some chart, moving along the loop by marking the tip of a unit vector at each point given by this orientation and connecting the two ends of this path once we loop around $\gamma$ once, we can create a loop $\eta$ intersecting $M$ at a single point $p$. Equivalently, this is a section of the normal bundle that vanishes at a single point.
Since $\mathbb{R}^{n+1}$ is contractible, the loop $\eta$ must be the boundary of a disc $D$ in $\mathbb{R}^{n+1}$. We can assume that $\partial D = \eta$ and $M$ are transverse. We can also assume that $D$ and $M$ are transverse after a perturbation that fixes $\partial D$. Hence the intersection $K = D\cap M$ is a compact $1$-dimensional manifold, possibly with boundary. Indeed, $\partial K = \partial(D\cap M) = \partial D \cap M$, thus its boundary is non-empty. But then $K$ must be a union of paths. In particular, its boundary contains at least $2$ points but $\partial D \cap M = \eta \cap M = \{p\}$. This is a contradiction.
In the above argument, I proved orientability because it seems like you observed that the original statement is equivalent to orientability. If it is not clear why this is the case and you can't figure out why, feel free leave a comment.
A: Here is a sketch of another proof, using Jordan-Brouwer's Theorem, which relies on the fact that $M$ is closed (that you did not explicitly mentioned in your question, but this is needed to apply this Theorem).
The other is already very good.
Let $A$ be the bounded component  of $\Bbb R^{n+1}\setminus M$.
Consider the signed distance $r\colon \Bbb R^{n+1}\to \Bbb R$ defined as
$$
r(x) =
\begin{cases}
-d(x,M) & \text{if } x \in A,\\
d(x,M) & \text{if } x \notin A.
\end{cases}
$$
Then:

*

*Show that $r$ is smooth in a neighbourhood of $M$,

*Show that $\nabla r |_{M}$ is an outward unit normal vector field on $M$.

