(a) The closure of a disconnected set is disconnected?
(b) What about the interior of a disconnected set?
For (a) take $\mathbb{R}^{2}-\{x\text{-axis}\}$.
For (b) seems true. I just had an intuitive idea. If a set $S$ is disconnected, $S = A \cup B$ where $A\cap B = \emptyset$, $\overline{A}\cap B = \emptyset$ and $A \cap \overline{B} = \emptyset$. So, take the interior of $S$ is take the interior of $A$ and the interior of $B$, that is, $\text{int }S \subset\text{int }A \cup \text{int }B$ and satisfies the hypothesis for disconnectedness.
But I dont know if its correct. Can someone help me?
(a) is correct, but an even simpler example (in the reals) is $(0,1) \cup (1,2)$, which has closure $[0,2]$.
For (b) look no further than the reals again: the disconnected set $(0,1) \cup \{2\}$ has interior $(0,1)$, quite connected.