Connected Subspaces of a Topological Space

Question

A few months ago I proved that if $A $ and $B$ are connected subspaces of a topological space $X$ and $A \cap B \neq \emptyset$ then $A \cup B$ is a connected subspace of $ X$. I did a proof by contradiction: Assume $A \cup B$ is disconnected. Then there exists open sets $C,D$, which are subsets of $A \cup B$, such that $C$ and $D$ are non-empty, their intersection is empty, and $C \cup D = A \cup B$. Since $A \cap B \neq \emptyset$ then there exists a $p$ belonging to $A$ and $B$. So $p$ is in $A \cup B$. $p$ is also in $C \cup D$. WLOG, let $p$ be in $C$. Since $p$ is in $C$ and $p$ is in $A$, then $p $ is in $A \cup C$. I cannot for the life of me remember why $A \cap C \neq \emptyset$ and $A$ being connected means $A$ is a subset of $ C$.

Answer 1

If $p$ is in $A$ and $C$ like you said, then $p\in A\cap C$, so $A\cap C\neq\varnothing$.

For your second question, let $A$ be a connected subspace of some of disconnected space $X$ (in the context of your problem, $X=A\cup B$). Then, $X=C\cup D$, where $C\cap D=\varnothing$ and are open in the topology of $X$. Suppose $A$ has elements in both $C$ and $D$. Because we suppose $A$ to be connected, we know that $A$ as a subspace endowed with the subspace topology must be connected. But because $A$ has elements in $C$ and $D$, $A\cap C$ and $A\cap D$ disconnect $A$ in the subspace topology, contradicting our claim that $A$ is connected in the subspace topology, which in turn contradicts our assumption that $A$ is connected in $X$. Thus, $A$ must be completely contained in either $C$ or $D$.

There is also a slightly shorter method to your proof using just the open sets (not specific points in $A$). Suppose $A$ and $B$ are connected in $X$, and $A\cup B$ is disconnected. Then, there exist $C,D\in(A\cup B,\tau_S)$ such that $A\cup B=C\cup D$, $C\cap D=\varnothing$, and $C,D\neq A\cup B$. But the only way for this to happen (with the condition that $A,B$ are connected in $X$ and by the result proved in the previous paragraph) is if $A$ is entirely contained in $C$ or $D$, and $B$ is entirely contained in the other set (e.g., $A\subseteq C$ and $B\subseteq D$). But because $C\cap D=\varnothing$, $A\cap B=\varnothing$. Thus, $A\cup B$ is disconnected in the subspace topology, which by definition makes it disconnected in $X$. Thus, if $A,B$ are connected in $X$ with $A\cap B\neq\varnothing$, then $A\cup B$ is also connected in $X$.