Recall the following theorem:
Theorem
Let $\{A_\alpha\}$ be a collection of connected subsets of $X$ such that $\bigcap_\alpha A_\alpha\neq\emptyset$. Then $\bigcup_\alpha A_\alpha$ is a connected subset of $X$.
I solved the following exercise (Exercise 2 page 152 of Munkres' Topology):
Exercise
Let $\{A_n\}$ be a sequence of connected subsets of $X$ such that $$ \forall n\in\mathbb{N},A_n\cap A_{n+1}\ne\emptyset. $$ Show that $\bigcup_{n\in\mathbb{N}}A_n$ is a connected subset of $X$.
After that, I attempted to generalize the result using transfinite induction. Could you please verify if my proof is correct, or tell me if there are any improvements? Thank you in advance.
Generalization
For any nonzero ordinal $\alpha$, we have the following:
"Let $\{A_\beta\mid\beta\in\alpha\}$ be a collection of connected subsets of $X$. Suppose that $$ \forall\beta\in\alpha,A_\beta\cap A_{\beta^+}\ne\emptyset $$ and that for any limit ordinal $\lambda\in\alpha$, there exists $\beta\in\lambda$ such that $A_\lambda\cap A_\beta\ne\emptyset$. Then $\bigcup_{\beta\in\alpha}A_\beta$ is a connected subset of $X$."
Proof. For any $\beta\in\alpha$, define
$$ B_\beta=\bigcup_{\gamma\underline{\in}\beta}A_\gamma $$
We prove using transfinite that all the $B_\beta$ are connected. Let $\beta\in\alpha$ and assume that for every $\gamma\in\beta$, $B_\gamma$ is connected.
$\beta=0$: $B_0=A_0$ is connected.
$\beta=\gamma^+$: We have $$ B_\beta=B_\gamma\cup A_{\gamma^+} $$ is the union of two connected sets and $$ \left(\bigcup_{\delta\underline{\in}\gamma}A_\delta\right)\cap A_{\gamma^+}\supset A_\gamma\cap A_{\gamma^+}\ne\emptyset. $$ Thus according to the theorem, $B_\beta$ is connected.
$\beta$ is a limit ordinal: We have $$ B_\beta=\left(\bigcup_{\gamma\in\beta}B_\gamma\right)\cup A_\beta. $$ Using the theorem, $\bigcup_{\gamma\in\beta}B_\gamma$ is connected. Indeed every $B_\gamma$ is connected and \begin{align*} \bigcap_{\gamma\in\beta}B_\gamma&=B_0\\ &=A_0\\ &\supset A_0\cap A_1\\ &\ne\emptyset. \end{align*}
Therefore for every $\beta\in\alpha$, $B_\beta$ is connected.
Now $\bigcup_{\beta\in\alpha}A_\beta=\bigcup_{\beta\in\alpha}B_\beta$ where all the $B_\beta$ are connected for $\beta\in\alpha$ and $\bigcap_{\beta\in\alpha}B_\beta=B_0\neq\emptyset$. The theorem ensures the desired result.$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\square$
Thank you for your remarks in comments. Here's the new generalization and proof:
Generalization
For any nonzero ordinal $\alpha$, we have the following:
"Let $\{A_\beta\mid\beta\in\alpha\}$ be a collection of connected subsets of $X$. Suppose that $$ \forall\beta\in\alpha,\exists\gamma\in\beta,A_\gamma\cap A_\beta\ne\emptyset. $$ Then $\bigcup_{\beta\in\alpha}A_\beta$ is a connected subset of $X$."
Proof. For any $\beta\in\alpha$, define
$$ B_\beta=\bigcup_{\gamma\underline{\in}\beta}A_\gamma $$
We prove using transfinite that all the $B_\beta$ are connected. Let $\beta\in\alpha$ and assume that for every $\gamma\in\beta$, $B_\gamma$ is connected.
$\beta=0$: $B_0=A_0$ is connected.
$\beta\neq 0$: We have $$ B_\beta=\left(\bigcup_{\gamma\in\beta}B_\gamma\right)\cup A_\beta. $$ Using the theorem, $\bigcup_{\gamma\in\beta}B_\gamma$ is connected. Indeed every $B_\gamma$ is connected and \begin{align*} \bigcap_{\gamma\in\beta}B_\gamma&=B_0\\ &=A_0\\ &\supset A_0\cap A_1\\ &\ne\emptyset. \end{align*}
By assumption, there exists $\gamma\in\beta$ such that $A_\beta\cap A_\gamma\ne\emptyset$. In other words, $$ A_\beta\cap\left(\bigcup_{\gamma\in\beta}A_\gamma\right)\ne\emptyset. $$ Since $\bigcup_{\gamma\in\beta}A_\gamma=\bigcup_{\gamma\in\beta}B_\gamma$, $B_\beta$ is connected according to the theorem.
Therefore for every $\beta\in\alpha$, $B_\beta$ is connected.
Now $\bigcup_{\beta\in\alpha}A_\beta=\bigcup_{\beta\in\alpha}B_\beta$ where all the $B_\beta$ are connected for $\beta\in\alpha$ and $\bigcap_{\beta\in\alpha}B_\beta=B_0\neq\emptyset$. The theorem ensures the desired result.$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\square$