26.14 Definition. A simple chain connecting two points $a$ and $b$ of a space $X$ is a sequence $U_1,\ldots,U_n$ of open sets of $X$ such that $a\in U_1$ only, $b\in U_n$ only, and $U_i\cap U_k\neq\varnothing$ iff $|i-j|\leqslant1$.
2.6.15 Theorem. If $X$ is connected and $\scr U$ is any open cover of $X$, then any two points $a$ and $b$ of $X$ can be connected by a simple chain consisting of elements of $\scr U$.
Proof. Let $Z$ be the set of all points of $X$ which are connected to $a$ by a simple chain of elements of $\scr U$. Then $Z$ is obviously an open st and, since $a\in Z$, $Z$ is nonempty. We can prove the theorem by showing $Z$ is closed.
$\quad$ Let $z\in\bar Z$. Then $z\in U$ for some $U\in\scr U$ and, $\color{brown}{\text{since $U$ is open, $U\cap Z$ contains }}$ $\color{brown}{\text{some point $b$}}$. Now $a$ is connected to $b$ by a simple chain $U_1,\ldots,U_n$ of elements of $\scr U$. If $z\in U_k$ for some $k$, then the smallest such $k$ produces a simple chain $U_1,\ldots,U_k$ from $a$ to $z$. If $z\notin U_k$ for any $k$, pick the smallest $l$ such that $U_1\cap U\neq\varnothing$ (e.g., $n$ is such an $l$). Then $U_1,\ldots,U_l$, $U$ is a simple chain from $a$ to $z$. Either way, $z\in Z.\;\blacksquare$
From Willard S. General Topology.
I have trouble to understand why, in the $\rm\color{brown}{brown}$ part, the intersection can't be empty.
can anyone help?
$\overline{Z}$ is the closure of $Z$, i.e the smallest closed set that contains $Z$ as a subset, or equivalently (one can show) the set of all $x$, such that every open set $U$ that contains $x$, intersects $Z$. So the statement follows from that last definition of the closure.