The set of matrices whose powers converge is neither open nor closed

The family. Let $D(x)=\operatorname{diag}(x,0,\dots,0)$. Then $D(x)^k=\operatorname{diag}(x^k,0,\dots,0)$, and $x^k$ converges iff $x\in(-1,1]$ (to $0$ for $|x|<1$, to $1$ for $x=1$; for $x=-1$ it oscillates, for $|x|>1$ it diverges). Thus
$$D(x)\in\mathcal{C}\iff x\in(-1,1].$$

Not open. $D(1)\in\mathcal{C}$, but $D(1+\varepsilon)\notin\mathcal{C}$ for every $\varepsilon>0$ and $D(1+\varepsilon)\to D(1)$. So $D(1)$ is not an interior point.

Not closed. $D(-1+\tfrac1j)\in\mathcal{C}$ for all $j\ge2$, and $D(-1+\tfrac1j)\to D(-1)$, but $D(-1)\notin\mathcal{C}$ since $(-1)^k$ has no limit. So $\mathcal{C}$ does not contain its limit points.

Remark. The interior of $\mathcal{C}$ is $\{\rho(A)<1\}$ (where $A^k\to0$) and its closure is the set of power-bounded matrices... more precisely, $\mathcal{C}$ is contained in $\{\rho\le1\}$ and contains $\{\rho<1\}$: the interesting matrices are on the boundary $\rho=1$, where convergence is governed by the Jordan structure of eigenvalue $1$ (semisimple) and the absence of other eigenvalues on the unit circle.