An example of a closed cone whose image under a linear map is not closed? I need a very simple example of a closed cone $K$ in $\mathbb{R}^n$ such that $A(K)$ is not closed where $A:\mathbb{R}^n \rightarrow \mathbb{R}^m$ is a linear map. 
The simplest ones I've come across are where $A$ is from $\mathbb{R}^3$ to $\mathbb{R}^2$, but is there one from $\mathbb{R}^2$ to $\mathbb{R}^1$? An example of $\mathbb{R}^3$ to $\mathbb{R}^2$ is this: the projection of the proper cone $K=\{(x,y,z):y^2\leq xz,z\geq 0\}$ onto the $(x,y)$-plane is not a proper cone (not proper because the projection is not closed).
(Definition of cone $K$: if $x\in K$ then $\lambda x\in K, \lambda \geq 0$ and $x+y\in K$ for all $x,y \in K$.) 
 A: There is no example for $\Bbb R^d\to\Bbb R,d\in\Bbb N$. 
Proof.
Let $K$ be a closed (convex) cones in $\Bbb R^d$ and $A:\Bbb R^d\to\Bbb R$ a linear map. Let $x_n\in AK\subset\Bbb R$ be a convergent sequence in $AK$. In order to show that $AK$ is closed, we have to prove that $\lim x_n\in AK$.
If $x_n\to0$, then the limit is obviously already in $AK$, since $0=A0\in AK$. Hence we assume that $x_n\to x\not=0$. Therefore, we can assume that w.l.o.g. all $x_n\in\Bbb R$ are non-zero and have the same sign. Especially, this implies $x_n/x_m>0$ for all $n,m\in\Bbb N$. 
Choose some non-zero $y_0\in K$ with $Ay_0=x_0$ and define $y_n:=y_0\cdot x_n/x_0$. Since, $x_n/x_0>0$, we have $y_n\in K$. Next, we see
$$\|y_n-y_m\|=\left\|y_0\frac{x_n}{x_0}-y_0\frac{x_m}{x_0}\right\|=\frac{\|y_0\|}{|x_0|}\cdot|x_n-x_m|.$$
So, because $x_n$ is a Cauchy sequence, $y_n$ must be one too. This means $y_n\to  y$ for some $y\in K$. Note that 
$$Ay_n=A\left(y_0\cdot \frac{x_n}{x_0}\right)=\frac{x_n}{x_0}\cdot Ay_0=\frac{x_n}{x_0}\cdot x_0=x_n$$
Since $A$ is continuous, we have that $\lim x_n=\lim (Ay_n)=A(\lim y_n)=Ay\in AK$. So, since any convergent sequence in $AK$ has its limit in $AK$ too, we conclude that $AK$ is a closed subset of $\Bbb R$. $\square$

I really love your example for where this fails for $\Bbb R^3\to\Bbb R^2$. The reason why above proof does not work for higher dimensional co-domains is, that there is no such way to "lift" the sequence $x_n\in AK$ to the cone $K$ while preserving convergence, as we did above with
$$y_n:=y_0\cdot\frac{x_n}{x_0}.$$
A: There cannot be a counterexample with $A \colon \mathbb R^d \to \mathbb R^1$:
$AK$
is a convex cone in $\mathbb R^1$ and there are only 4 different convex cones in $\mathbb R^1$: $\{0\}$, $[0,∞)$, $(−∞,0]$ and $\mathbb R^1$. All of them are closed. 
Moreover, there cannot be a counterexample with $A \colon \mathbb R^2 \to \mathbb R^2$: If $A$ has full rank, it is invertible and thus $AK$ is closed (it is the preimage of $K$ w.r.t. $A^{-1}$). If $A$ does not have full rank, the situation is essentially one-dimensional and the above objection applies.
Hence, the minimal counterexample has $A \colon \mathbb R^3 \to \mathbb R^2$ and you have found it :)
