About a limit-set which consists of a single point in a compact topological space Let $X$ be a compact topological space. For $S\subset X$, $\mathrm{cl}\,\{S\} $ denote the closure of $S$ in $X$.
Denote by  $\mathrm{lp}(x_{n}):=\bigcap_{n\in\mathbb{N}} \{\mathrm{cl}\, \{x_{m}:m\ge n\} \}$, where $\{x_{n}\}$ is a sequence in $X$.
Why $\mathrm{lp}(x_{n})=\{x\}$ implies that $\{x_{n}\}$ must converges to $x$ ?
 A: Let $U$ be an open neighborhood of $x$, and let $A_k = \text{cl}\{x_m \mid m \geq k\}$. We want to show that for any open neighborhood $U$, it contains all but finitely many terms in the sequence $\{x_n\}$, i.e. there is some $M$ such that $x_m \in U$ for all $m \geq M$. Since we are given compactness, we want to create an open cover of something and use that it admits a finite subcover. What first comes to mind is to use the $A_k$, which are closed, and take their complements to form open sets. Then
$$\bigcup_{k \in \mathbb{N}} X \setminus A_k = X \setminus \bigcap_{k \in \mathbb{N}} A_k = X \setminus \{x\} \supseteq X \setminus U$$
Here we used that $x \in U$, noting that if $X$ contains sets $A,B$ then $A \subseteq B$ if and only if $X \setminus A \supseteq X \setminus B$.
Since each $A_k$ is closed, $X \setminus A_k$ is open so this is an open cover for $X \setminus U$. Since $U$ is open, $X \setminus U$ is closed, and hence compact as closed subsets of compact sets are compact. Thus, this admits a finite subcover (compact means every open cover admits a finite subcover)
$$\bigcup_{i=1}^N X \setminus A_{k_i} = X \setminus \bigcap_{i=1}^N A_{k_i}\supseteq X \setminus U$$
Using the same thing as above, we have shown that
$$\bigcap_{i=1}^N A_{k_i} \subseteq U$$
Do you see where to go from here? Use the definition of the $A_k$'s
