How to prove each point of X is an isolated point of BX and no other point of BX-X is an isolated point in BX??
Since closure of X is BX so there cannot exist any isolated point. Am I correct?
I Cannot prove each isolated point of X is an isolated point of BX. Please help
If $p$ is isolated in $X$ then $\{p\}$ and $X\setminus \{p\}$ are disjoint closed sets of $X$ and so they have disjoint closures in $\beta X$. As $\beta X$ is $T_1$, the former closure is still $\{p\}$ and $\beta X = \overline{X} = \{p\} \cup \overline{X\setminus \{p\}}$ (closures in $\beta X$) and so $\{p\}$ is also open in $\beta X$ as the complement of $\overline{X\setminus\{p\}}$.
Another argument: as $\{p\}$ is open in $X$, there is an open set $O$ of $\beta X$ such that $O\cap X=\{p\}$. This means that $O=\{p\}$ as otherwise $O \setminus \{p\}$ would be a non-empty open subset of $\beta X$ that misses the dense set $X$, which cannot be. This only uses the denseness of $X$ and $\beta X$ being $T_1$.
If $\{p\}$ were an isolated point point of $\beta X$ in with $p \notin X$, then $\{p\}$ is an open set of $\beta X$ missing the dense subset $X$, which cannot be.