Compact Metric Space with sequence

Question

$(1)$ If $S$ is a compact metric space, and $\{x_{n} \}_{n \in \mathbb{N}} \subset S$ is a sequence with finitely many accumulation points, $y_{1}, \dots ,y_{k}$, then we can partition $\mathbb{N} = \cup_{j=1}^{k} S_{j} $, such that $ \{ x_{n} : n \in S_{j} \}$ converges to $y_{j}$, $\forall j \in [k]$.
$(2)$ If $S$ is an arbitrary metric space and $\{x_{n} \}$ has a countably infinite amount of accumulation points, $y_{1}, y_{2} ,\dots $, then we can partition $\mathbb{N} = \cup_{j=1}^{\infty} S_{j}$ such that $\{ x_{n} : n \in S_{j} \}$ converges to $y_{j}$, $\forall j \in \mathbb{N}$.
For the first question, since $S$ is sequentially compact, we have that $\exists \{x_{n_{k}} \} \subset \{x_{n} \}$ such that $x_{n_{k}} \rightarrow y_{1}$. Then $S_{1} = \{n_{k} : k \in \mathbb{N} \}$. Then repeat the procedure on the sequence $X_{2} = \{ x_{n} : n \in \mathbb{N} \setminus S_{1} \}$. It has a subsequence which converges by sequential compactness to $y_{2}$, and denote this index set as $S_{2}$. Then define $X_{t} = \{x_{n} : n \in \mathbb{N} \setminus \cup_{j=1}^{t} S_{j} \}$, and it has a subsequence which converges to $y_{t}$. Does this procedure work and if so why?
For the arbitrary metric space, I am not sure how to proceed.

Answer 1

No, your procedure does not work, because there is no guarantee that you will pick up every natural number, and hence you won't get a partition.

Instead, here's a way to proceed for (1). Let $r > 0$ be the minimal distance between pairs of points $y_1,...,y_k$. Let $\epsilon = r/3$. Consider the disjoint union of balls $U = B(y_1,\epsilon) \cup \cdots \cup B(y_k,\epsilon)$. Argue that there exists a natural number $N$ such that if $n \ge N$ then $x_n \in U$; if not then, applying sequential compactness, you'll get a subsequence that does not converge to any of $y_1,...,y_k$. You can now partition the infinite ray $\{N,N+1,N+2,N+3,...\}$ into $k$ sets, according to which of those balls contains $x_n$ for each $n \ge N$. The rest of the indices $\{1,...,N-1\}$ can be assigned to partition elements arbitrarily.