Let $X$ be a topological space. Prove that the following are equivalent: (a) $X$ is connected. (b) For every collection $\{U_\alpha\}$ of open subsets of $X$ with $X = \cup_\alpha U$ and every two points $x, y ∈ X$, there are finitely many $U_1, U_2, . . . , U_n \in \{U_\alpha\}$ such that $x ∈ U_1$, $U_i \cap U_{i+1} \neq \emptyset$ for all $1 \le i < n$, and $y ∈ U_n$.
I am really confused with this given information, what does for all x belong to X, x belong to U1 mean??
This equivalence condition for connectedness, says that any pair of distinct points $x,y$ of $X$, one can conjoin them with a chain of open sets. Remember that in path connectedness, we conjoin two points with a path.
For a scketch of the proof, let $X$ be connected and consider the open cover $\{U_\alpha\}$ of $X$. Suppose $A\subset X$ to be the set of points of $X$ which can conjoin together with a finite chain of open sets. Obviously $A\ne\varnothing$ since any point can conjoin to itself. Prove that $A$ is both open and closed. Then connectedness of $X$ implies $ A=X $.
For the converse, let $f:X\to\{0,1\}$ be continuous. The property of ability to conjoin any two pair of distinct points of $X$ implies that $f$ compel to be constant. So $X$ is connected.