Let $X$ be a $US$ space. Then $X^*$ is $US$ iff in $X$ , every convergent sequence has a relatively compact subsequence $( X^*,\tau^*)$ is one - point compatification of topological space $ (X, \tau)$.
A topological space is called a $US$-space provided that
each convergent sequence has a unique limit.
The bellow theorem is  in " Between T_{1} and T_{2}" by " Wilansky"
Theorem: Let $X$ be a $US$ space. Then $X^*$ is $US$ iff in $X$ , every convergent sequence has a relatively compact subsequence.
Prrof: if $X^*$ is $US $ , and‎ $ x_{n} \longrightarrow  a$ in $X$ , then $x_{n} \not\longrightarrow  \infty$ in $X^\infty *$. Thus there is an $\tau^*$- neighborhood $G$ of$\infty $ s.t $x_{k(n)}  \not\in G $ for some sequence $ \{k(n)\}$ . since the complement of $G$ is a compact closed of $X$ , the closure of $x_{k(n)}$ is compact.
Conversely, let the condition hold and let $ \{ x_{n} \}$ be a sequence in 
$X^*$ with $ x_{n} \longrightarrow  a \not =\infty  $. Let $K$ be the closure ( in $X$) of a relatively subsequence. Then $X^* - K$ is a  $\tau^*$-neighborhood of $\infty $ and it is false that $x_{n} $ belongs to eventhally . Thus  $ x_{n} \not\longrightarrow \infty  $.

(1) Why he said that "since the complement of $G$ is a compact closed of $X$ , the closure of $x_{k(n)}$ is compact" ?
(2) Why he said that "Then $X^* - K$ is a  $\tau^*$-neighborhood of $\infty $ and it is false that $x_{n} $ belongs to eventhally . Thus  $ x_{n} \not\longrightarrow \infty  $?

 A: Once again there are quite a few errors in your copying from the paper.


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*$X\setminus G=X^*\setminus G$ is compact and closed in $X$ by the definition of $X^*$: open nbhds of $\infty$ are the complements of compact, closed subsets of $X$. Let $A=\{x_{n_k}:k\in\omega\}$; $A\cap G=\varnothing$, so $\operatorname{cl}_XA$ is a $\tau$closed subset of the $\tau$-compact set $X\setminus G$ and is therefore $\tau$-compact. (You keep forgetting: closed subsets of compact sets are always closed.)

*Once again, this is just the definition of $X^*$: $K$ is a $\tau$-compact subset of $X$, so by definition $X^*\setminus K$ is a $\tau^*$-open nbhd of $\infty$ in $X^*$. Let $\langle x_{n_k}:k\in\omega\rangle$ be the subsequence such that $K$ is the closure of $\{x_{n_k}:k\in\omega\}$. Then for each $k\in\omega$ we have $x_{n_k}\in K$ and hence $x_{n_k}\notin X^*\setminus K$. Clearly this implies that there is no $m\in\omega$ such that $x_n\in X^*\setminus K$ for all $n\ge m$, so $\langle x_n:n\in\omega\rangle$ does not converge to $\infty$.
