$X^{\ast }$ is compact and contains $X$ as an open subspace; $X$ is dense in $X^{\ast }$ exactly when $X$ is not compact; and $X^{\ast }$ is Hausdorff exactly when $X$ is locally compact and Hausdorff

Claim 2: $X\in \mathcal{T}$, so $X\in {\mathcal{T}}^{\ast }$ by [L1] and $X$ is open in $X^{\ast }$; and the traces on $X$ of the members of ${\mathcal{T}}^{\ast }$ are the sets $U\cap X=U$ for $U\in \mathcal{T}$ and the sets $(X^{\ast }\setminus C)\cap X=X\setminus C$ for $C$ closed in $X$, all of which lie in $\mathcal{T}$, while every $U\in \mathcal{T}$ is its own trace. So the subspace topology is $\mathcal{T}$.

Claim 1: let $\mathcal{U}\subseteq {\mathcal{T}}^{\ast }$ have union $X^{\ast }$; some $O\in \mathcal{U}$ contains $\mathrm{\infty }$, so $O=X^{\ast }\setminus C$ with $C$ closed in $X$ and a compact subset of $X$, by [L1].

For claim 3, the open subsets of $X^{\ast }$ containing $\mathrm{\infty }$ are exactly the sets $X^{\ast }\setminus C$ with $C\subseteq X$ closed and compact, and $(X^{\ast }\setminus C)\cap X=X\setminus C$; so by [L5] the point $\mathrm{\infty }$ lies in the closure of $X$ exactly when $X\setminus C\mathrm{\ne }\mathrm{\varnothing }$ for every such $C$.

For the backward half of claim 4 assume $X$ is locally compact and Hausdorff, and let $u\mathrm{\ne }v$ in $X^{\ast }$. If both lie in $X$, disjoint open subsets of $X$ separating them are open in $X^{\ast }$ by [L1]. If $v=\mathrm{\infty }$ and $u=x\in X$, then [L7] applied to the neighbourhood $X$ of $x$ gives a compact neighbourhood $C$ of $x$, closed by [L6], and an open $U$ of $X$ with $x\in U\subseteq C$; then $U$ and $X^{\ast }\setminus C$ are disjoint members of ${\mathcal{T}}^{\ast }$ containing $x$ and $\mathrm{\infty }$.

For the forward half of claim 4 assume $X^{\ast }$ is Hausdorff. Distinct points of $X$ are separated in $X^{\ast }$ by disjoint open $P,Q$, and $P\cap X$, $Q\cap X$ are disjoint sets open in $X$ by claim 2, so $X$ is Hausdorff.

The traces $W\cap X$ for $W\in \mathcal{U}$ are open in $X$ by step 1.1, and they cover $C$, since $C\subseteq X$ and $\bigcup \mathcal{U}=X^{\ast }$; so [L3], applied with index set $\mathcal{U}$ and the family $W\mapsto W\cap X$, gives $n\in \mathbb{N}$ and $W_0,\dots ,W_n\in \mathcal{U}$ with $C\subseteq (W_0\cap X)\cup \cdots \cup (W_n\cap X)$, or else $C=\mathrm{\varnothing }$.

A closed compact $C\subseteq X$ equals $X$ exactly when $X$ is compact, since $X$ is closed in $X$ and, by [L2], $X$ is a compact subset of itself exactly when it is a compact space. So the condition of step 1.3 fails for some $C$ exactly when $X$ is compact.

For $x\in X$ the Hausdorff property of $X^{\ast }$ gives disjoint open $U\ni x$ and $O\ni \mathrm{\infty }$; by [L1] $O=X^{\ast }\setminus C$ with $C$ closed in $X$ and compact, and $U\cap O=\mathrm{\varnothing }$ forces $U\subseteq X^{\ast }\setminus O=C$. As $U$ is open in $X$ by step 1.1 and contains $x$, the compact set $C$ is a neighbourhood of $x$ in $X$ by [L4], so $X$ is locally compact.

Claim 1 follows: $X^{\ast }=O\cup W_0\cup \cdots \cup W_n$, since a point of $X^{\ast }$ is either $\mathrm{\infty }$ or a point of $X$, a point of $X$ outside $C$ lies in $O=X^{\ast }\setminus C$, and a point of $C$ lies in some $W_j$ by step 2.2; in the alternative $C=\mathrm{\varnothing }$ already $X^{\ast }=O$. So every open cover of $X^{\ast }$ has a finite subcover.

Claim 3 follows: $\bar{X}=X^{\ast }$ holds exactly when $\mathrm{\infty }\in \bar{X}$, since $X\subseteq \bar{X}$ and $X^{\ast }=X\cup \{\mathrm{\infty }\}$; by steps 1.3 and 2.3 that holds exactly when $X$ is not compact.

Claims 1, 2, 3 and 4 are established: claim 1 at step 3.2, claim 2 at step 1.1, claim 3 at step 3.3, and claim 4 by steps 1.4 for one direction and 2.1 and 3.1 for the other.