Connectedness of a set covered by connected sets

Question

Suppose that we can write $X = \bigcup_{i=1\cdots n} A_i$, where each $A_i$ is connected and for each $i = 2 \cdots n, A_i \cap A_{i-1} \neq \emptyset$ (means that consecutive $A_k$ are not disjoint). Prove that $(X, τ_X)$ is also connected.

The proof I was given starts like this :

Let $U$ and $V$ be two disjoint open subsets of $X$ such that $X = U \sqcup V$. Assume also that $U$ is non-empty. Note that since for each $i \in I, A_i$ is connected, either $A_i ⊆ U$ or $A_i ⊆ V$.

The last sentence (in bold) is not clear to me, it visually feels true but it is not proven rigorously here. What if $X$ is some subset of $\Bbb R$ and $U$ is some interval like $(\cdots,\frac12)$ and $V=[\frac12,\cdots)$ such that $X = U \sqcup V$, then if an $A_i$ is for example $(\frac12-\epsilon,\frac12+\epsilon)$ for some $\epsilon > 0$, we have that $A_i$ is connected but we do not have that $A_i ⊆ U$ or $A_i ⊆ V$.

Can someone explain why is the bold written sentence true ?

Answer 1

We have $A_i=(A_i\cap U)\sqcup (A_i\cap V)$ and these sets are open in the subspace topology of $A_i$.

Now using the definition of connectedness, we conclude that either $A_i\cap U$ or $A_i\cap V$ is empty and the statement follows.

Answer 2

I think I found something by myself :

Since $A_i$ is connected, if $A_i= X_i \cup Y_i$ with $X_i, Y_i$ disjoint open sets, then in the case that $X_i \subset U$ and $Y_i \subset V$ either $X_i=A_i$ or $Y_i=A_i$.