Sketch $M=\{re^{i\phi}:r\in(1,2],\phi\in[0,\pi)\}$ and give its closure $\overline{M}$. Is $M$ compact?
$\overline{M} = \{re^{i\phi}:r \in (1,2], \phi \in [0,\pi) \} \cup \{ re^{i\pi} \} \cup \{e^{i\phi}\} \cup \{ e^{i\pi} \}$.
$M$ is not compact as for example $[0, e^{i\pi}]\subseteq M^c$.
Is my understanding of this correct? If yes, does it need any more argumentation?
Your picture is not correct. Since $\theta\in[0,\pi)$, it should be just the top half (also with $(1,2]$, but without $[-2,1)$).
The set $M$ is not closed because, for instance $1\in\overline M\setminus M$. But $\overline M$ is compact, since it is both closed and bounded.