Can a sequence have infinitely many subsequences that are disjoint?

I wanted to prove the following,

Let $\{x\_n\}$ be a sequence of a metric space with no converging subsequences. Then for each point outside the sequence $\{x\_n\}$ has an open neighborhood that has no intersection with $\{x\_n\}$.

Here is how I proceeded.

Let $\{ y\_n \}$ be a subsequence and let $x$ be a point not in $x\_n$. Since $y\_n$ is not a convergent sequence it does not converge to $x$.
Therefore, for all $N>0$, $\exists \epsilon>0$ such that $\{y\_{n>N}\} \cap B(x,\epsilon) =
\emptyset$.
Since metric space is $T\_0$ space, for any point $y\_k \in \{y\_{n}\} $ we can find $B(x,\epsilon\_k)$ not containing $y\_k$. Now the set $ U\_y= \cap\_{k\le N} B(x,\epsilon\_k) \cap B(x,\epsilon) $ is an open set(intersection of finite number of open sets) with $U\_y \cap \{y\_n\} =\emptyset$.

Now consider the set $\{x\_n\} - \{y\_n\}$ and let $\{z\_n\}$ be a subsequence of $\{x\_n\} - \{y\_n\}$. Then can find an open set $U\_z$ such that $U\_z \cap \{z\_n\} =\emptyset$. Assume $\{x\_n\}$ can only have a finite number of disjoint subsequences.

Then continue like that until we find finite number of points $S=\{x\_n\} - \{y\_n\}-\{z\_n\} ...$. Again using the $T\_0$ property, for each point $s \in S$, we can find an open ball around $x$, $B(x,\epsilon\_s)$ that does not contain the point $s$. Therefore The set $U= \cap\_{s\in S} B(x,\epsilon\_s) \cap U\_y \cap U\_z \cap .. $ is a finite intersection of open sets that contain $x$ but does not contain $\{x\_n\}$.

For the proof, I needed to use that there are only finitely many disjoint subsequences in $\{x\_n\} $which I think should not be an assumption. Any comments or thoughts are really appreciated.