Q:

$T_1$ property and local compactness

Recall that a space  $X$ is a locally compact if for every $x\in X$ there exists a neighbourhood $U$ of the point of $x$ such that $cl U$  is compact subspace of $X$. How we can show that  compact $cl U$ is $T_1$?

A:

$\newcommand{\cl}{\operatorname{cl}}$Engelking defines compact to mean what I would call compact and Hausdorff. It is true that if each point of a space $X$ has a nbhd $U$ such that $\cl U$ is compact and Hausdorff, then $X$ is $T_1$. More generally, if $X$ is any space, and $x\in X$ has a nbhd $U$ such that $\cl U$ is compact and Hausdorff, then $\{x\}$ is closed in $X$. (Recall that a space is $T_1$ iff each singleton set is closed.)
To see this, let $x\in X$, and let $U$ be a nbhd of $x$ such that $\cl U$ is compact and Hausdorff. The space $\cl U$ with the relative topology inherited from $X$ is a Hausdorff space, so it’s $T_1$, and therefore $\{x\}$ is a closed set in $\cl U$. This means that there is some closed set $F$ in $X$ such that $\{x\}=F\cap\cl U$. But then $\{x\}$ is the intersection of two sets that are closed in $X$, and as such it must itself be closed in $X$. 

