Suppose that $A$ and $B$ are connected subsets of $X$ that are not separate from each other (either $A \cap \overline{B}$ is non-empty, $\overline{A} \cap B$ is non-empty or both are non-empty). Prove that $A \cup B$ is also connected.
In Baby Rudin, a set was said to be connected if it was not the union of two non-empty separated sets. How does that relate to the notion of separation in this case? And how will one proceed with this question in general?
I like the following characterisation of connected spaces
A metric (topological) space $X$ is connected iff every continuous map $f: X \to \{0,1\}$ where $\{0,1\}$ has the usual metric topology (which is discrete) is constant.
Thus, to show that $A \cup B$ is connected, consider a continuous map $f: A \cup B \to \{0,1\}$. Then the restrictions $f\vert_A: A \to \{0,1\}$ and $f\vert_B: B \to \{0,1\}$ are also continuous, and thus constant. Since $A \cap B \neq \emptyset$, you know that these both maps must be constant with the same value. Thus $f$ is constant on $A \cup B$.
We conclude that $A \cup B$ is connected.
Suppose that $A\cap\overline B$ is connected. Let $C$ be a subset of $A\cap\overline B$ which is both open in $A\cap\overline B$ and closed in $A\cap\overline B$ and take $p\in A\cap\overline B$. Then $p\in C$ or $p\in V^\complement$. If $p\in C$, then, since $C\cap A$ is non-empty, open in $A$ and closed in $A$, $C\cap A=A$. In other words, $C\supset A$. By the same argument, and since $\overline B$ is connected, $C\supset\overline B$. So, $C=A\cap B$.
