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

Claim 2: $X \in \mathcal{T}$, so $X \in \mathcal{T}^{*}$ by [L1] and $X$ is open in $X^{*}$; and the traces on $X$ of the members of $\mathcal{T}^{*}$ are the sets $U \cap X = U$ for $U \in \mathcal{T}$ and the sets $(X^{*} \setminus C) \cap X = X \setminus C$ 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 $\mathcal{U} \subseteq \mathcal{T}^{*}$ have union $X^{*}$; some $O \in \mathcal{U}$ contains $\infty$, so $O = X^{*} \setminus C$ with $C$ closed in $X$ and a compact subset of $X$, by [L1].

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

For the backward half of claim 4 assume $X$ is locally compact and Hausdorff, and let $u \ne v$ in $X^{*}$. If both lie in $X$, disjoint open subsets of $X$ separating them are open in $X^{*}$ by [L1]. If $v = \infty$ 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 $X^{*} \setminus C$ are disjoint members of $\mathcal{T}^{*}$ containing $x$ and $\infty$.

For the forward half of claim 4 assume $X^{*}$ is Hausdorff. Distinct points of $X$ are separated in $X^{*}$ 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 $\bigcup \mathcal{U} = X^{*}$; 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 $C \subseteq (W_0 \cap X) \cup \dots \cup (W_n \cap X)$, or else $C = \varnothing$.

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 $X^{*}$ gives disjoint open $U \ni x$ and $O \ni \infty$; by [L1] $O = X^{*} \setminus C$ with $C$ closed in $X$ and compact, and $U \cap O = \varnothing$ forces $U \subseteq X^{*} \setminus O = C$. 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: $X^{*} = O \cup W_0 \cup \dots \cup W_n$, since a point of $X^{*}$ is either $\infty$ or a point of $X$, a point of $X$ outside $C$ lies in $O = X^{*} \setminus C$, and a point of $C$ lies in some $W_j$ by step 2.2; in the alternative $C = \varnothing$ already $X^{*} = O$. So every open cover of $X^{*}$ has a finite subcover.

Claim 3 follows: $\overline{X} = X^{*}$ holds exactly when $\infty \in \overline{X}$, since $X \subseteq \overline{X}$ and $X^{*} = X \cup \{\infty\}$; 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.