Can every bounded sequence be partioned into a set of convergent subsequences? This thought was prompted by thinking about a proof of the Boundedness Theorem for a real analysis module in the second year of a Mathematics BSc course.
\begin{align}
& \text{Let $(x_n)$ be a bounded sequence.} && (1) \\
& \text{Then $(x_n)$ has a convergent subsequence $(x_{n_k})$, by the Bolzano-Weierstrass Theorem.}  && (2) \\
& \text{Now let $(y_n)=(x_n)\backslash(x_{n_k})$.}  && (3) \\
& \text{Then $(y_n$) is a bounded sequence and so has a convergent subsequence $(y_{n_k})$.}  && (4)
\end{align}
Repeating this process, don't we partition $(x_n)$ into convergent subsequences?
 A: I agree with Ethan Bolker. I vote for him.
But note that you will not necessarily obtain every limit point of your original sequence as a limit of one of your subsequences. So I think that undermines the usefulness.
For example, consider the sequence $a_n = \sin(n)$ for $n \in \{1,2,\dots\}$. Then every point $x \in [-1,1]$ is a limit point of some subsequence. You could get one sequence $\mathfrak{X}^{(1)} = (x^{(1)}_1,x^{(1)}_2,x^{(1)}_3,\dots)$ converging to say $0$, such that $x^{(1)}_1=\sin(1)$. Then you can get a second sequence and a third sequence $\mathfrak{X}^{(i)} =(x^{(i)}_1,x^{(i)}_2,x^{(i)}_3,\dots)$ for $i=2,3$ converging to say $1/2$ and $-1/2$ with $x^{(2)}_1=\sin(k_2)$ where $k_2$ is the smallest $k$ such that $\sin(k)$ is not in $\mathfrak{X}^{(1)}$ (which is necessarily at least $2$) and $x^{(3)}_1=\sin(k_3)$ where $k$ is the smallest $k$ such that $\sin(k)$ is not in $\mathfrak{X}^{(1)}$ or $\mathfrak{X}^{(2)}$ (which is necessarily at least $3$). You could continue in this way to get subsequences converging to all the numbers $0$ and $\pm i/2^n$ for $n \in \{1,2,3,\dots\}$ and $i \in \{1,\dots,2^{n-1}\}$. But you will not be able to get all numbers in $[-1,1]$ as limits of these subsequences.
Actually, you could also get a different set of subsequences $\mathcal{Y}^{(1)}$, $\mathcal{Y}^{(2)}$, $\dots$ converging only to the numbers $1/2^n$ for $n \in \{1,2,\dots\}$ because you can use up all the negative values of $\sin(n)$ (as well as all the values greater than $1/2$) as first elements of new subsequences.
