Every cluster point of a bounded sequence $(x_n)$ is the $\mathcal{F}$-limit for some free filter $\mathcal{F}$ Let $c$ be a cluster point of a bounded sequence $(x_n)$. I am trying to prove that there exists a free filter $\mathcal{F}$ s.t. 
$$c= \lim_\mathcal{F} \ (x_n) .$$

My attempt: Since $c$ is a cluster point of $(x_n)$, for every neighbourhood $V$ of $c$ and every $k \in \mathbb{N}$, 
$$\{n \geq k : x_n \in V \} \neq \emptyset$$ 
It is enough to define
$$ \mathcal{F}:= \big\{ \{n \geq k : x_n \in V \}     \big\}_{k \in \mathbb{N}, \ V \in \mathcal{V}(c)}$$
because it is a free filter and $c= \lim_\mathcal{F} \ (x_n): $ 
Free filter:
1.- Since $\{n \geq k : x_n \in V \} \neq \emptyset$, $\emptyset \notin \mathcal{F}$.
2.- $\{n \geq k_1 : x_n \in V_1 \} \cap \{n \geq k_2 : x_n \in V_2 \} = \{n \geq \max \{ k_1,k_2\} : x_n \in V_1 \cap V_2 \} \neq \emptyset $
3.- Is every $A$ s.t. $\{n \geq k : x_n \in V \} \subset A \subset \mathbb{N}$ also in $\mathcal{F}$? I am not sure of it, maybe not but $\mathcal{F}$ is basis... .
4.- $ \cap \mathcal{F} = \emptyset$ as $k \in \mathbb{N}$. No $n$ can be in every set!
For $c= \lim_\mathcal{F} \ (x_n) $ it is enough to take $k=1$, so $\{n  : x_n \in V \} \in \mathcal{F}$ $\blacksquare$

Question is: is this proof correct? What about 3.- ?
 A: Corrected. Let $C=\{n\in\Bbb N:x_n=c\}$. Your approach works fine if $C=\varnothing$, and you can prove $(3)$ as follows. 
For each $V\in\mathscr{V}(c)$ and $k\in\Bbb N$ let $F_V(k)=\{n\ge k:x_n\in V\}$. Suppose that $A\supseteq F_V(k)\in\mathscr{F}$. Let $U=V\cup\{x_n:n\in A\}$; then $U\supseteq V\in\mathscr{V}$, so $U\in\mathscr{V}$, and $A=F_U(0)\in\mathscr{F}$.
If $C$ is infinite, let $\mathscr{F}=\{F\subseteq\Bbb N:C\setminus F\text{ is finite}\}$; it’s not hard to show that $\mathscr{F}$ is a free filter on $\Bbb N$ such that the $\mathscr{F}$-limit of $\langle x_n:n\in\Bbb N\rangle$ is $c$, since $\{n\in\Bbb N:x_n\in V\}\supseteq C$ for each $V\in\mathscr{V}$.
If $C$ is finite but non-empty, it seems easiest first to construct a filter on $\Bbb N\setminus C$ as in the second paragraph: let $M=\Bbb N\setminus C$, for each $V\in\mathscr{V}(c)$ and $k\in\Bbb N$ let 
$$F_V(k)=\{n\in M:n\ge k\text{ and }x_n\in V\}\;,$$
and let $\mathscr{F}=\{F_V(k):k\in\Bbb N\text{ and }V\in\mathscr{V}\}$. Then, as in the case $C=\varnothing$, $\mathscr{F}$ is a free filter on $M$, and $c$ is the $\mathscr{F}$-limit of $\langle x_n:n\in M\rangle$. $\mathscr{F}$ is a base for the free filter 
$$\mathscr{G}=\{F\cup S:F\in\mathscr{F}\text{ and }S\subseteq C\}$$
on $\Bbb N$, and $c$ is the $\mathscr{G}$-limit of $\langle x_n:n\in\Bbb N\rangle$.
