Suppose both $A_1 \cup A_2$ and $A_1 \cap A_2$ are connected. Show that
a) If $A_1$ and $A_2$ are closed then $A_1$ and $A_2$ are connected.
b) if one of $A_1$ or $A_2$ is not closed, then $A_1$ or $A_2$ may be disconnected.
Here is what I got so far.
Supposed that both $A_1 \cup A_2$ and $A_1 \cap A_2$ are connected, and $A_1$ and $A_2$ are closed. Assume the contrary that $A_1 $ is disconnected. Then there exist 2 non-empty disjoint open set $B_1$ and $C_1$ such that $$A_1 \subset B_1 \cup C_1$$
Thus $$A_1 \cup A_2=B_1 \cup C_1 \cup A_2$$
since $B_1 \cap C_1 =\emptyset$, $A_1 \cap A_2= A_2 \cap B_1 \cup A_2 \cap C_1$ is disconnected, contradict to the assumption that it's connected. We cna use similar argument for $A_2$ . the thing that concern me is I don't use any from the fact that $A_1$ and $A_2$ are closed. So I got the feeling that my argument is not correct. For part b), I'm not sure where to start, should I find an example?
To show (a), argue by contraposition. Without the loss of generality, assume that $A_1, A_2$ are closed and $A_1$ is disconnected. Here using the fact that $A_1$ is closed, there has to exist an open interval with positive length $(a,b) \subset {A_1}^c$ such that the end points $a$ and $b$ are contained in $A_1$.
Now if $A_1\cup A_2$ is connected, then $(a,b)\subset A_2$. Using the fact that $A_2$ is also closed, the end points $a,b\in A_2$ as well. However $A_1\cap A_2$ would be disconnected since $a,b\in A_1\cap A_2$ but $(A_1\cap A_2)\cap(a,b) = \emptyset$.
Thus $A_1\cup A_2$ and $A_1\cap A_2$ can not be both connected.