Is each closed convex set a manifold with corners? Assume that $C$ is a convex set in $\mathbb{R}^{n}$ with non empty interior.
Then consider its closure, is it a smooth manifold with corners?
Edit: 
1) The closure of $C$ should be a smooth manifold with corners equipped with the induced smooth structure from $\mathbb{R}^{n}$.
2) With non empty interior I mean that $C$ contains a non empty open set.
 A: Consider the following curve (very informally described):


*

*start from the origin in $\mathbb{R}^2$, then move from one unit up.

*Turn of an angle $\pi/4$ on the left and move from half of unit.

*Turn of an angle $\pi/8$ on the left and move from a quarter of a unit.

*...

*Turn of an angle $\pi/2^{n+2}$ on the left and move from $1/2^{n+1}$ unit.

*...


If you parametrised this using an interval of length one for each step you get a function from $\mathbb{R}$ to $\mathbb{R}^2$ whose derivative is always in the upper left quarter of the plane and which converge to a point $p=(x,y)$ in the upper left quarter of the plane. Add the segments $(x,t)$ with $t$ between $0$ and $y$ and $(v,0)$ with $v$ between $x$ and $0$.

The inside (in the sense of Jordan's theorem) of this curve is a closed convex subset of $R^2$ of non-empty interior which cannot be endowed with a structure of differentiable manifold with corner which makes the maps to $R^2$ differentiable: Indeed such a structure would have to treat all the angles in the curve as corner, and because they have an accumulation point being itself a corner this is not going to be possible: the point $p$ has no neighbourhood diffeomorphic to an open of $[0,\infty[^k$, any of its neighbourhood contains an infinite number of corners.
Of course if you are talking about topological manifold or if you don't want your manifold structure to be compatible with the differentiable structure induced from $\mathbb{R}^2$ then this is no longer a counter example: the things I defined is homeomorphic to the closed disk so there is a structure of $C^{\infty}$ manifold with boundaries compatible to its topology, so that is why I said that the question need to be made more precise.
A: See de Rham's "cutting corners" curve.  



Picture from an MO answer by Bill Thurston; also see a description there.
The limiting curve is $C^1$ but not $C^2$.  It has a tangent everywhere, but curvature zero almost everywhere.  When you say "manifold with boundary" what do you require?
Plug: English translation of de Rham's paper is in my book Classics on Fractals
A: Let $C \subset S^1 \subset \mathbb{R}^2$ be a Cantor set. Let $H_C$ be its convex hull, the smallest closed subset of $\mathbb{R}^2$ containing $C$. Then $H_C$ is not a manifold with corners. Its boundary $\partial H_C$ is $C^1$ at every point except for the countably many points which are endpoints of components of $S^1-C$, these points forming a dense subset of $C$. Also, $\partial H_C$ fails to be $C^2$ at every point of $C$. But $H_C$ is homeomorphic to the closed 2-disc, indeed there is an ambient homeomorphism $\mathbb{R}^2 \mapsto \mathbb{R}^2$ taking $H_C$ to the unit closed disc.
