Convergence of subsequences in a compact set of $\mathbb{R}^n$ For any sequence in a compact set $K$ of $\mathbb{R}^n$, it is well known that there is a subsequence that is convergent to a point in $K$.
Is this true that for any sequence in a compact set $K$ of $\mathbb{R}^n$, there is a partition of this sequence into several subsequences that every subsequence is convergent to a point in $K$. 
Intuitively, it is correct to me. Any constructive proof?

Denote the original sequence by $\{a_n\}$ and call its convergent sub-sequence $\{a_{n_0}\}$
Construct $\{a_{n_1}\}$ by first removing the terms of $\{a_{n_0}\}$ from $\{a_n\}$ and then finding a convergent sub-sequence.
Proceeding in this way, each time you will construct $\{a_{n_k}\}$ by first removing the terms of $\{a_{n_{k-1}}\}$ from $\{a_{n_{k-2}}\}$ and then finding a convergent sub-sequence of what remains, which is possible because of compactness. Then $\{a_{n_k}\}$'s form a partition of $\{a_n\}$, not necessarily a finite partition, such that each $\{a_{n_k}\}$ is a convergent sequence.
Q.E.D.
Addendum: One might wonder if one could always find a finite partitioning of the original sequence. To see that finite partitioning isn't enough, consider this example:
Take a bunch of sequences $b_i=\{ a_{i,n} \}_{n\in \mathbb{N}}$ such that $\lim_{n\to \infty}b_i=l_i$ where $l_i$'s are different.
Now write down each sequence $b_i$ in a row to form an infinite square. Create a new sequence $\{c_n\}$ by covering this square by first going down the first column, then going up the second column, then again going down the third column and so forth. The sequence that is obtained in this way will have no finite partitioning.
