Let $\{A_n\}$ be a sequence of connected subspaces of $X$, such that $A_n\cap A_{n+1}\neq \emptyset$ for all $n$. Show that $\bigcup A_n$ Is connected.
My attempt:
Approach(1): Assume towards contradiction, $\bigcup A_n$ is not connected. Let $P$ and $Q$ form a separation of $\bigcup A_n$. $\forall n\in \Bbb{N}, \exists p_n\in A_n \cap A_{n+1}\neq \emptyset$. So $p_n\in P$ or $p_n \in Q$. Claim: $p_n \in P$, $\forall n\in \Bbb{N}$ or $p_n\in Q$, $\forall n\in \Bbb{N}$. Proof: Assume towards contradiction, $\exists a,b\in \Bbb{N}$ with $a\lt b$(WLOG) such that $p_a\in P$ and $p_b\in Q$. By exercise 1 section 16, $A_n$ is a subspace of $\bigcup A_n$, $\forall n\in \Bbb{N}$. By lemma 23.2, $A_n\subseteq P$ or $A_n\subseteq Q$, $\forall n\in \Bbb{N}$. Since $p_a\in A_a \cap A_{a+1}$, we have $A_a ,A_{a+1} \subseteq P$. $\exists k\in \Bbb{N}$ such that $a+k=b$. $A_{a+1}\cap A_{a+2}\neq \emptyset$, $\exists p_{a+1}\in A_{a+1}\cap A_{a+2}$. Since $p_{a+1}\in A_{a+1}\subseteq P$, $A_{a+2}\subseteq P$. Repeating this process $k$ times. So $A_{a+k}=A_b \subseteq P$. Which contradicts our initial assumption of $p_b\in Q$ (or $A_b\subseteq Q$). Thus our assumption must be wrong. Hence $p_n \in P$, $\forall n\in \Bbb{N}$ or $p_n\in Q$, $\forall n\in \Bbb{N}$. WLOG assume $p_n\in P$, $\forall n$. Let $n\in \Bbb{N}$. $A_n$ and $A_{n+1}$ are connected and subspace of $\bigcup A_n$. Since $p_n\in P$, we have $A_n,A_{n+1} \subseteq P$. So $A_n \cup A_{n+1}\subseteq P$. Since $n$ was arbitrary, $A_n \cup A_{n+1}\subseteq P$, $\forall n\in \Bbb{N}$. Thus $\bigcup_{n\in \Bbb{N}} A_n \subseteq P$. Which implies $Q=\emptyset$. Which contradicts the fact that $Q\neq \emptyset$. Is this proof correct? This proof is inspired by Munkres’ proof of theorem 23.3. Can this approach is refined(especially claim part)?
Approach(2): Let $f:\bigcup A_n \to \{0,1\}$ be a continuous map. $f|_{A_n}: A_n \to \{0,1\}$ defined by $f|_{A_n}(x)=f(x),\forall x\in A_n$ is a continuous map by theorem 18.2. Since $A_n$ is connected, $f|_{A_n} (A_n)= f(A_n)=\{i_n\}$, for some $i_n\in \{0,1\}$. Since $A_n \cap A_{n+1}\neq \emptyset$, $\exists x_n\in A_n \cap A_{n+1}$. So $f(x_n)=i_n=i_{n+1}, \forall n\in \Bbb{N}$. Let $i_1=i$. Claim: $i_n =i, \forall n\in \Bbb{N}$. Proof: Base case, $i_1=i$ is true. Suppose $i_k=i$ for some $k\in \Bbb{N}$. Then $i_k =i_{k+1}=i$. So $i_{k+1}=i$. By mathematical induction $i_n=i, \forall n\in\Bbb{N}$. Thus $f(\bigcup A_n)=i\in \{0,1\}$, a constant map. Hence $\bigcup A_n$ is connected. Is this proof correct?
Approach(3): https://dbfin.com/topology/munkres/chapter-3/section-23-connected-spaces/problem-2-solution/.
Approach(4): $\{A_n\}$ is a sequence of connected subspace of $X$ s.t. $A_n \cap A_{n+1} \not= \emptyset$ for all $n$. Show that $\cup A_n$ is connected or similar proof.
Application: Inspired by https://math.stackexchange.com/a/4385310/861687 proof.
Claim: If $\{ A_n| n\in \Bbb{N}\}$, $A_n$ connected subspaces of $X$ such that $A_i\cap A_j\neq \emptyset, \forall i,j\in \Bbb{N}$, then $\bigcup_{n\in \Bbb{N}} A_n$ is connected.
Proof: Condition $A_n\cap A_{n+1}\neq \emptyset ,\forall n\in \Bbb{N}$ is weaker then $A_i\cap A_j\neq \emptyset, \forall i,j\in \Bbb{N}$ , i.e. $A_i\cap A_j\neq \emptyset, \forall i,j\in \Bbb{N} \Rightarrow A_n\cap A_{n+1}\neq \emptyset ,\forall n\in \Bbb{N}$. By exercise 2 section 23, $\bigcup A_n$ is connected.
