A problem from Spivak's Calculus on Manifolds Notation
As Spivak suggests, given $A\subset\mathbb R^n$, boundary $A$ denotes the topological boundary of $A$, i.e. $\overline A\cap\overline{A^c}$.

Problem 5-3(a): Let $A\subset\mathbb R^n$ be an open set such that boundary $A$ is an $n-1$ dimensional manifold. Show that $N=A\cup$ boundary $A$ is an $n$ dimensional manifold-with-boundary.

Thoughts
It seems that we only need to show that there's a neighborhood $V_x$ of $x\in$ boundary $A$ such that $V_x\cap N$ is diffeomorphic to $\mathbb R^n$ or $[0,\infty)\times \mathbb R^{n-1}$. The first situation is met when $A=\{x:0<\lVert x\rVert<1\}$.
The condition of boundary $A$ is indispensable, but I don't know how to make use of it. At first, I found a counterexample, eventually pointed out by somebody that it is against the condition of boundary $A$:
$$A=\bigcup_{n=1}^\infty\left(\frac1{2^{n+1}},\frac1{2^n}\right)\times\mathbb R$$
It seems that, the material of manifolds is not that manageable.
 A: You want to use everything given to you - in particular, you've ignored the definition of $\partial A$ as the set of limit points of $A.$ The proof follows below.
By assumption, for any $x\in\partial A,$ there exist open sets $x\in U \subset \mathbb{R}^{n},\, V\subset \mathbb{R}^{n},$ and a diffeomorphism $g: U \to V$  such that $g(U\cap \partial A) = V\cap \mathbb{R}^{n-1}\times\{0\}.$ What we want to show is that $g(U'\cap (A \cup \partial A)) = V'\cap (\mathbb{R}^{n-1}\times[0,\infty))$ or $V'\cap \mathbb{R}^{n}$ for some open subsets $U'\ni x$ of $U$ and $V'$ of $V.$
Consider a sequence of points $A\ni\{x_k\}\to x$ for which $\{g(x_k)\}\to g(x)$ lies entirely in (one of) the half plane(s) $\mathbb{R}^{n}_{>0} = \mathbb{R}^{n-1}\times (0,\infty)$ (or $\mathbb{R}^{n-1}\times(-\infty,0)$). Now, suppose that there is no open neighborhood $V'$ of $g(x)$ such that $\mathbb{R}^{n}_{>0}\cap V' \subset g(U).$ This means that there is a sequence of points in $R^{n}_{>0} - g(U)$ that converges to $g(x).$ Let us call this sequence $\{y_k\}.$ Then it must be the case that the line segment connecting $g(x_k)$ and $y_k$ (for each $k$) contains a point in $g(U\cap\partial A),$ but this is obviously impossible since $g(U\cap \partial A)$ includes only those points with last coordinate $0.$ 
Remark: Note that it is not necessarily true that $\partial A = \partial N.$ 
