Let $\{A_\alpha \}$ be a collection of connected subspaces of $X$; let $A$ be a connected subspace of $X$. Show that if $A\cap A_\alpha \neq \emptyset$ for all $\alpha$, then $A\cup (\bigcup A_\alpha)$ is connected.
My attempt:
Approach(1): let $f: A\cup (\bigcup A_\alpha)\to \{0,1\}$ be a continuous map. $f|_{A_\alpha}: A_\alpha \to \{0,1\}$ and $f|_A:A\to \{0,1\}$ is a continuous map by theorem 18.2. Since $A\cap A_\alpha \neq \emptyset$, $\exists p_\alpha \in A\cap A_\alpha$, $\forall \alpha$. Since $A_\alpha$ and $A$ is connected, $f|_{A_\alpha} (A_\alpha)= f(A_\alpha)=\{i_\alpha \}=\{ f(p_\alpha)\}$ and $f|_A(A)=f(A)=\{ i\}=\{f(p_\alpha)\}$; $p_\alpha \in A\cap A_\alpha$, $\forall \alpha \in J$. So $i=f(p_\alpha)=i_\alpha ,\forall \alpha \in J$. Thus $i=i_\alpha$, $\forall \alpha \in J$. Hence $f$ is constant and $A\cup (\bigcup A_\alpha)$ is connected. Is this proof correct?
Approach(2): https://math.stackexchange.com/a/703142/861687. $\bigcap (A\cup A_\alpha)=A \cup (\bigcap A_\alpha)\neq \emptyset$, since $A\neq \emptyset$. IMO this proof is more clever and slick.