Show that the closed $n$-ball $B^n(a)$ is a manifold.
I know how to show that $S^{n-1}(a)=\partial B^n(a)$ is an $n-1$ manifold without boundary. We consider the function $f(x)=a^2-\Vert x\Vert ^2$. And the open ball $B^n(a)$ is trivially an $n$-dimensional manifold, since it is open in $\mathbb R^n$. But how about the closed ball? I cannot use the same function as for $S^{n-1}(a)$, because $\det Df(0)=0$, so we have a problem there. Any ideas on how to fix this?
To do this, you need to show that every point $x \in B^n(a)$ has a neighborhood $U$ which is homeomorphic to an open subset of $\mathbb{H}^n = \{(x_1, \ldots, x_n) \ | \ x_n \geq 0\}$. This is the upper half space, and has boundary $\{(x_1, \ldots, x_n) \ | \ x_n = 0\}$. The points $x$ which are mapped to the boundary of $\mathbb{H}^n$ by the above-described homeomorphism form the boundary of the manifold.
Since clearly each point $x$ in the interior of $B^n(a)$ has a neighborhood which is homeomorphic to an open subset of $\mathbb{H}^n$ (just by, say, translating a small ball around $x$ upwards so that now it lies in the upper half plane).
So the only issue is showing that each point on the boundary $S^{n - 1}$ has an open neighborhood homeomorphic to an open neighborhood of $\mathbb{H}^n$. This can be done by "straightening the boundary" of the sphere. To do this, try using stereographic projection (see https://www.physicsforums.com/threads/closed-ball-is-manifold-with-boundary.744620/).
It would probably help to do this in 2 dimensions first; the generalization step should be almost identical.