Boundary of the set of points away from manifold is a hypersurface (This is part of Problem 6-6 from Lee's Introduction to Smooth Manifolds textbook.)

Suppose $M\subset \mathbb{R}^n$ is a compact embedded submanifold. For
  any $\epsilon>0$, let $M_\epsilon$ be the set of all points in
  $\mathbb{R}^n$ whose distance from $M$ is less than $\epsilon$. Show
  that for sufficiently small $\epsilon$, $\partial M_\epsilon$ is a
  compact embedded hypersurface in $\mathbb{R}^n$.

I thought to use the tubular neighbourhood theorem (since we have an embedded submanifold) and showing for some sufficiently small tubular neighbourhood, maybe we can prove that the boundary of the tubular neighbourhood is an embedded $(n-1)$-dimensional embedded manifold (and relate $\epsilon$ to the distance function in the definition of a tubular neighbourhood). However, I have no idea how to prove that (or if it's even the right approach).
For reference, I'm using Lee's definition of a tubular neighbourhood. That is, $E(V)$ where $E(x,v)=x+v$ and $V=\{(x,v)\in NM : |v|<\delta(x)\}$ for positive continuous $\delta: M \rightarrow \mathbb{R}$. 
 A: As you suggest, using the tubular neighborhood theorem is the right idea. 
Summary. The core point is as follows: Let $y \in \mathbb{R}^n$ have $d(y,M) = \ell$. Then since $M$ is compact there exists a point $x \in M$ with $d(y,x) = \ell$, and the line segment from $x$ to $y$ in $\mathbb{R}^n$ intersects $M$ orthogonally at $x$. (This last fact is a multivariable calculus exercise: The gradient of $d(\cdot, y)$ is orthogonal to $M$ at any minimum.) This lets you relate "distance from $M$" to a condition on the normal bundle of $M$, which lets you use tubular neighborhoods.
I'll flesh out some details without being too careful, but I think everything below is fairly rote given the above:
Review of stuff in Lee. Some reminders so that we're on the same page: Let $M \subseteq \mathbb{R}^n$ be an embedded submanifold. The points of the normal bundle $NM$ of $M$ are pairs $(x, v)$ with $x \in M$ and $v \in T_{x}\mathbb{R}^n$ normal to $M$. There's a map $E : NM \to \mathbb{R}^n$ given by $E(x,v) = x+v$. 
Suppose $f: M \to \mathbb{R}$ is a positive continuous function on $M$ and let $V$ be the subset of $NM$ given by
$$
  V = \lbrace (x, v) \in NM : |v| < f(x) \rbrace.
$$
If $U = E(V)$ is open and $E$ restricts to a diffeomorphism $V \cong U$, then $U$ is called a tubular neighborhood of $M$ in $\mathbb{R}^n$. Lee proves (it is Theorem 10.19 in my copy) that every embedded $M \subseteq \mathbb{R}^n$ has a tubular neighborhood.
Notation. For any $\delta > 0$ we define $V_\delta \subset NM$ by
$$
  V_\delta = \lbrace (x, v) \in NM : |v| < \delta \rbrace.
$$
When $E$ gives a diffeomorphism of $V_\delta$ onto an open subset of $\mathbb{R}^n$, we denote by $U_\delta$ the image of $V_\delta$.
When $M$ is compact, any $f$ giving a tubular neighborhood has a positive minimum, and so there exists a constant $\delta > 0$ such that $M$ has a tubular neighborhood of the form $U_\delta = E(V_\delta)$. 
Let $\epsilon < \delta$. 
The Main Point. $M_\epsilon$ is contained within $U_\delta = E(V_\delta)$. This follows from the discussion in our summary at the top; if $x$ is a point in $M$ closest to $y$, then $y$ is the image under $E$ of $(x, y-x)$, which is in $V_\delta$.
In fact, this argument shows that $M_\epsilon = U_\epsilon$.
It follows that $\partial M_\epsilon$ is the image under $E$ of the set
$$
  \partial V_\epsilon = \lbrace (x, v) \in NM : |v| = \epsilon \rbrace.
$$
Since $\partial V_\epsilon$ is clearly an embedded submanifold of $V_\delta$, and $E$ is a diffeomorphism from $V_\delta$ onto $U_\delta$, it follows that $\partial M_\epsilon$ is an embedded submanifold of $U_\delta$ (and hence of $\mathbb{R}^n$).
Similarly, since $\partial V_\epsilon$ has codimension $1$, we see that $\partial M_\epsilon$ does as well. 
(In fact this shows that $\partial M_\epsilon$ is the diffeomorpic to the bundle of unit $(n-m)$-spheres in $NM$; we can picture $\partial M_\epsilon$ as a "hollow tube" around $M$ whose "cross-section" at any $x \in M$ is an $(n-m)$-sphere. It follows easily from this that $\partial M_\epsilon$ is compact.)  
