Here is Prob. 2, Sec. 23, in the book Topology by James R. Munkres, 2nd edition:
Let $\left\{ A_n \right\}$ be a sequence of connected subspaces of (a topological ) space $X$, such that $A_n \cap A_{n+1} \neq \emptyset$ for all $n$. Show that $\bigcup A_n$ is connected.
I think I'm able to make sense of this proof.
Indeed, I assert that the above result could be generalised as follows:
Let $X$ be any topological space; let $J$ be an arbitrary non-empty, well-ordered set; let $\left\{ A_\alpha \right\}_{\alpha \in J}$ be a collection of connected subspaces of $X$, such that, for each $\alpha \in J$, we have $$ A_\alpha \cap A_{s(\alpha)} \neq \emptyset, $$ where $s(\alpha)$ denotes the immediate successor of $\alpha$ in $J$; and let $$ A \colon= \bigcup_{\alpha \in J} A_\alpha. $$ Then $A$ is also connected.
Am I right?
If so, then here is my attempted proof of this result:
Suppose that $A$ is not connected, and let $U, V$ be a separation of $A$. For each $\alpha \in J$, as $A_\alpha$ is connected, so $A_\alpha$ lies entirely within either $C$ or $D$ (but not both), by Lemma 23.2 in Munkres.
Let $\alpha_0$ denote the smallest element of the well-ordered set $J$. We may assume without any loss of generality that $$ A_{\alpha_0} \subset C. $$
Let $J_0$ denote the following subset of $J$: $$ J_0 \colon= \left\{ \ \alpha \in J \ \colon \ A_\alpha \subset C \ \right\}. $$ Then by our assumption $\alpha_0 \in J_0$ so that $J_0$ is non-empty. We show that $J_0$ is inductive.
Let $\alpha \in J$ be arbitrary, and let us suppose that $S_\alpha \subset J_0$, where $S_\alpha$ denotes the section of $J$ by $\alpha$, that is, $$ S_\alpha \colon= \{ \ \xi \in J \colon \ \xi < \alpha \ \}. $$ We need to show hence that $\alpha \in J_0$ also.
How to show this?
Or, alternatively, if $J_0$ is a proper subset of $J$, then the set $J \setminus J_0$ is a non-empty subset of the well-ordered set $J$ and so has a smallest element $\beta_0$, say. Then $$ S_{\beta_0} \subset J_0. $$
How to proceed from here?
Here is a counterexample to your assertion with well-ordered sets.
Consider the well ordered set $\{ 0, 1 \} \times \mathbb{N}$ with the dictionary order: $(k_1,n_1) < (k_2,n_2)$ if and only if either $k_1 < k_2$, or $k_1=k_2$ and $n_1<n_2$.
Let $X \colon= \{ 0, 1 \}$ with the discrete topology.
Define $A_{(0,n)} = \{0\}$ and $A_{(1,n)} = \{1\}$. The union of all the $A$'s is not connected, but your hypothesis is satisfied, since the successor of $(k,n)$ is $(k,n+1)$ and $A_{(k,n)}=A_{(k, n+1)}=\{k\}$, for each $k \in \{ 0, 1 \}$.