Existence of a discrete family of sets Given that in a metric space $(X,d)$, there exists a sequence $(x_n)$ which does not cluster in $X$. Then how to construct a discrete family of sets $V_n$ such that each $V_n$ is a closed neighbourhood of $x_n$? By a discrete family of sets, we mean a family such that every point of $X$ has a neighbourhood which intersects at most one element of the family.
What is the corresponding result for general topological space?
Thank you

For $n\in\Bbb N$ let $r_n=\frac13\inf\{d(x_n,x_k):k\ne n\}$; since $x_n$ is not a cluster point of the sequence, $r_n>0$. Now for $n\in\Bbb N$ let $V_n$ be the closed ball of radius $\frac{r_n}{2^n}$ centred at $x_n$. I’ll leave it to you to check the details; the choice of $r_n$ ensures that the sets $V_n$ are pairwise disjoint, and the factor of $2^{-n}$ ensures that any limit point of $\bigcup_{n\in\Bbb N}V_n$ is a cluster point of the sequence.
In general topological spaces it may not be possible to find such nbhds. Let 
$$\begin{align*}
L&=\left\{\left\langle\frac1n,0\right\rangle:n\in\Bbb Z^+\right\}\;,\\
Y&=\left\{\left\langle\frac1n,\frac1m\right\rangle:m,n\in\Bbb Z^+\right\}\;,\text{ and}\\
p&=\langle 0,0\rangle\;,
\end{align*}$$
and let $X=\{p\}\cup L\cup Y$. Let $\tau_e$ be the Euclidean topology on $X$, and let $\tau$ be the topology generated by the base $\tau_e\cup\{U\setminus L:U\in\tau_e\}$. You can easily check that this is the usual topology on $L\cup Y$, and that a set $U\subseteq X$ is a nbhd of $p$ if and only if $p\in U$, and $U$ contains all but finitely many points of $\left\{\frac1n\right\}\times\left\{\frac1m:m\in\Bbb Z^+\right\}$ for all but finitely many $n\in\Bbb Z^+$.
Now let $x_n=\left\langle\frac1n,0,\right\rangle$ for $n\in\Bbb Z^+$. The sequence $\langle x_n:n\in\Bbb Z^+\rangle$ has no cluster point in $X$, but if $V_n$ is a closed nbhd of $x_n$ for each $n\in\Bbb Z^+$, every open nbhd of $p$ intersects infinitely many of the sets $V_n$.
