First of all we remember some elementary definitions and results about manifolds with corners.
Definition
A function $f$ defined in a subset $S$ of $\Bbb R^k$ is said of class $C^r$ if it can be extended to a function $\phi$ (said $C^r$-extension) that is of class $C^r$ in a open neighborhood of $S$.
Lemma
If $f$ is a function defined in a subset $S$ of $\Bbb R^n$ such that for any $x\in S$ there exists a function $f_x$ defined in a neighborhood of $x$ that is of class $C^r$ and compatible with $f$ on $U_x\cap S$ then $f$ is of class $C^r$.
Lemma
If $U$ is an open set of $H^n_k:=\Bbb R^{n-k}\times[0,+\infty)^k$ for any $k\le n$ then the derivatives of two different extensions $\phi$ and $\varphi$ of a $C^r$-function $f$ agree in $U$.
Definition
A $k$-manifold with corners in $\Bbb R^n$ of class $C^r$ is a subspace $M$ of $\Bbb R^n$ whose points have a neighborhood $V$ in $M$ that is the immage of a homeomorphism $\phi$ of calss $C^r$ defined an open set $U$ of $\Bbb R^k$ or of $H^k_m$ and whose derivative has rank $k$.
Definition
A point $y$ of a $k$-manifold with corners $M$ is said interior point or boundary point if there exist a coordinate patch about $y$ defined in an open set of $\Bbb R^k$ or in an open set of $H^k_l$ respectively.
Theorem
Let be $M$ a $k$-manifold with corners in $\Bbb R^n$. So if $\phi:U\rightarrow V$ is a coordinate patch about any point boundary point $y$ of $M$ then necessarily
$$
y=\phi(x)
$$
for any $x\in U\cap\operatorname{bd}H^k_l$.
So with the previous definitions and efforts I ask to explain (rigorously) why the boundary points set $\partial M$ of a $k$-manifolds with corners is or is not a $(k-1)$-manifolds with corners too: indeed at the page $253$ of the text Introduction to Smooth Manifolds by John M. Lee it is explicitely said that generally the boudary of a smooth manifolds with corners is not a smooth manifolds with corners although it is not said that it is not a (not smooth) manifold with corners and so this confunsed me. Moreover I do not know if my definition of manifold with corners is effectively compatible with John M. Lee definition although it seemed to me it is.
So provided that the result is true I arranged a proof that I show to follow: I point out that the proposed solution is imperfect because I did not able to prove a little but important thing as me myself I am showing. So could someone help me, please?
MY PROOF ATTEMPT
Well by the preceding theorem we know that if $y$ is an element of $\partial M$ then there exist a coordinate patch $\phi:U\rightarrow V$ defined in an open set of $H^k_l$ such that
$$
y=\phi(x)
$$
for any $x\in\operatorname{bd}H^k_l$ and thus without loss of generality we can assume that $x^k$ is zero: indeed if $x\in\operatorname{bd}H^k_l$ then it must be
$$
x^i=0
$$
for any $i=(k-l)+1,\dots,k$ so that if $\psi$ is the diffeomorphism that interchanges the $i$-th coordinate with the last (observe that $\psi$ is an involution that maps $H^k_l$ onto $H^k_l$) then $\phi\circ\psi$ is a coordinate patch about $y$ having the desired property. So we remember (click here for details) that if $W$ is an open set of $\Bbb R^k$ then
$$
W\cap\big(\Bbb R^{k-1}\times\{0\}\big)=A_W\times\{0\}
$$
for any open set $A_W$ of $\Bbb R^{k-1}$. Now we first assuem that it is $l=1$. So in this case the restriction of $\phi$ to $U\cap\operatorname{bd}H^k_1$ carries this set in a one to one fashion onto the open set $V\cap\partial M$ of $\partial M$. Now if $U$ is open in $H^k_1$ then there must exist an open set $W$ of $\Bbb R^k$ whose intersection with $H^k_1$ is equal to $U$ and thus putting
$$
\varphi(x):=\phi(x,0)
$$
for any $x\in A_W$ then we prove that $\varphi$ is a coordinate patch about $y$. So clearly $\varphi$ is of class $C^r$ because $\phi$ is; moreover the derivative of $\phi$ has rank $(k-1)$ because $D\varphi(x)$ consists simply of the first $(k-1)$ columns of the matrix $D\phi(x,0)$; finally $\phi$ is injective and its immage $V\cap\partial M$ is open in $M$ and the inverse function $\varphi^{-1}$ is continuous because it is equal to the function $\phi^{-1}$ followed by the projection of $\Bbb R^k$ onto its first $(k-1)$ coordinates. So we conclude that $\varphi$ is a coordinate patch about $y$ defined in an open set of $\Bbb R^{k-1}$. Now we let assume that $l>1$. So since $U$ is open in $H^k_l$ then
$$
U=W\cap H^k_l
$$
for any open set $W$ of $\Bbb R^k$ and thus
$$
U\cap\operatorname{bd}H^k_1=(W\cap H^k_l)\cap\operatorname{bd}H^k_1=(W\cap\operatorname{bd}H^k_1)\cap H^k_l=\\\big(W\cap(\Bbb R^{k-1}\times\{0\})\big)\cap(H^{k-1}_{l-1}\times[0,+\infty))=(A_W\times\{0\})\cap(H^{k-1}_{l-1}\times[0,+\infty))=(A_W\cap H^{k-1}_{l-1})\times\{0\}
$$
so that we let to prove that the function $\varphi$ defined in $A_W\cap H^{k-1}_{l-1}$ through the equation
$$
\varphi(x):=\phi(x,0)
$$
for any $x\in A_W\cap H^{k-1}_{l-1}$ is a coordinate patch about $y$. So clearly $A_W\cap H^{k-1}_{l-1}$ is open in $H^{k-1}_{l-1}$ and by analogous arguments applied above $\varphi$ is a $C^r$ injective function whose derivative has maximum rank and whose inverse is continuous but unfortunately I did not able to show that the immage of $\varphi$ is effectively open in $\partial M$ so that I can not claim that $\varphi$ is a coordinate patch about $y$.
Best Answer
Actually, every smooth manifold with corners is also a topological manifold with boundary, and therefore its boundary is a topological manifold (without boundary).
To see why, let $\overline{\mathbb R}{}^n_+$ be the "model corner space": $$ \overline{\mathbb R}{}^n_+ = \{ (x^1,\dots,x^n)\in\mathbb R^n: x^1\ge 0,\dots,x^n\ge 0\}. $$ It's relatively easy to show that $\overline{\mathbb R}{}^n_+$ is homeomorphic to the closed upper half-space $\mathbb H^n =\{(x^1,\dots,x^n): x^n\ge 0\}$ (see Exercise 16.18 in my ISM). Thus every smooth chart with corners can be composed with a homeomorphism $ \overline{\mathbb R}{}^n_+\to \mathbb H^n$ to obtain a topological boundary chart, and the result follows.