Let $(X,\mathscr T)$ be a topological space and let $a,b \in X$. A simple chain connecting $a$ and $b$ is a finite set $U_1,U_2,...,U_n$ of open sets such that $a\in U_1\setminus U_2$, $b\in U_n\setminus U_{n-1}$, and each $i,j=1,2,...,n, U_i\cap U_j \neq \emptyset$ iff $|i-j|\leq 1.$
Let $\mathscr O$ is an open cover of $X$. Then $A=\{x\in X:$ there is a simple chain consisting of members of $\mathscr O$ that connects $a$ and $x\}$ is open.
My attempt:- Let $x\in A$, I need to find $(x\in U)\in \mathscr T:U\subset A$. $x\in A$ means there is a simple chain consisting of members of $\mathscr O$ that connects $a$ and $x.$ That is there is a finite set $U_1,U_2,...,U_n\in \mathscr O$ of open sets such that $a\in U_1\setminus U_2$, $x\in U_n\setminus U_{n-1}$, and each $i,j=1,2,...,n, U_i\cap U_j \neq \emptyset$ iff $|i-j|\leq 1.$ How do I get $A$ is open from here? Please help me.
Let $x\in A$, and let $U_n$ be the last open set in a chain $U_1,...,U_n$ connecting $a$ and $x$.
Then $U_n$ is an open neighborhood of $x$.
To prove $A$ is open, it suffices to prove that $U_n\subseteq A$.
Let $y\in U_n$.
We want to show that $y\in A$.
If $n=1$, the chain $U_1$ connects $a$ and $y$, hence $y\in A$.
Suppose $n > 1$.
If $y\not\in U_{n-1}$, the chain $U_1,...,U_n$ connects $a$ and $y$, hence $y\in A$.
Suppose $y\in U_{n-1}$.
If $n=2$, the chain $U_1$ and connects $a$ and $y$, hence $y\in A$.
If $n > 2$, then since $y\in U_n$, it follows that $y\not\in U_{n-2}$ (since $U_n$ and $U_{n-2}$ are disjoint), hence the chain $U_1,...,U_{n-1}$ connects $a$ and $y$, so we get $y\in A$.
Thus, in all cases, we have $y\in A$.
It follows that $U_n\subseteq A$.
Therefore $A$ is open.