Infinite metric space has open set $U$ which is infinite and its complement is infinite 
Let $(X,d)$ be a metric space where $X$ is an infinite set. Prove that the space has an open set $U$ such that both $U$ and its complement are infinite sets.

I have considered if $d$ is the discrete metric, then one such $U$ exists. But if $d$ is not discrete, how do we find such $U$?
 A: If $p\in X$ and $p$ is not an isolated point of $X$ (that is,if $p\in \overline {X\backslash \{p\}}$) then every nbhd of $p$ is infinite.(Because we may take $p_1\in X\backslash \{p\}$, and take $p_{n+1}\in  B_d(p,(1/2)d(p,p_n))\backslash \{p\}$ for each $n\in N.$ Any nbhd of $p$ contains $p_n$ for all but finitely many $n.$) So if $X$ has two unequal non-isolated points $p, q$, then $U=B_d(p,(1/2)d(p,q))$ is open and infinite,and $U$ is disjoint from the infinite set $B_d(q,(1/2)d(p,q))$.On the other hand if $X$ has at most one non-isolated point, let $\{x_n :n\in N\}$  be a set of isolated points of $X$, with  $(n\ne m\to x_n\ne x_m).$ Then $U=\{x_{2n}:n\in N\}$ is open and infinite and $U$ is disjoint from the infinite set $\{x_{2n-1}:n\in N\}.$
A: The statement is correct, and I am trying to prove it by contradiction. Suppose that there exists an infinite metric space $(X,d)$ such that every infinite open set has finite complement.
Recall that a limit point of a metric space is a point whose open neighbourhoods are infinite.
Then it is quite easy to understand that:


*

*$X$ is not discrete, hence it has at most one limit point $p$

*$p$ is the unique limit point of $X$: otherwise, denoting by $p,p'$ two limit points, find $U, U'$ two disjoint open sets separating $p$ and $p'$, these are both infinite, against our assumptions on $X$.

*$X$ is compact: if $U_i$ is an open cover, one of these open sets is a neighbourhood of $p$ and its complement is finite.


An easy example of a metric space satisfying these three properties is
$$K= \{ 1/n \}_{n \ge 1} \cup \{ 0 \} \subset \Bbb{R}$$
Anyway, $K$ has an infinite open set with infinite complement, namely:
$$U= \{ 1/2n\}_{n \ge 1}$$
Following this behaviour of $K$, take any infinite set $A \subset X$ with infinite complement. Clearly the closure of $A$ is
$$\overline{A}=A \cup \{ p \}$$
 since $p$ is the unique limit point of $X$. This shows that the infinite open set $X \setminus \overline{A}$ has infinite complement: we reached a contradiction. Hence such a metric space does not exist.
