Bounded divergent sequence in $\mathbb{R}^n$ has subsequences congergent to at least two limits The problem that I'm presented with is as follows:
Let $(x_k)$ be a bounded sequence in $\mathbb{R}^n$, and 
let $a$ be a vector in $\mathbb{R}^n$.  It is given that $(x_k)$ does not 
converge to $a$.  (However, there is the possibility that there exist some subsequences of $(x_k)$ which converge to $a$.). Prove that there exists a vector $b \neq a$ in $\mathbb{R}^n$, and indices $1 \leq k(1) < k(2) < \cdots < k(p) < \cdots$ in $\mathbb{N}$, such that
$\lim\limits_{p \to \infty} x_{k(p)} = b$.
If my understanding is correct then what I need to prove is that even if there exists a subsequence of $(x_k)$ which converges to $a$ then there necessarily exists a vector $a\in \mathbb{R}^n$ such that at least one subsequence of $(x_k)$ also converges to $a$.
Here's my proof:
Since $(x_k)$ is a bounded sequence, $\exists m,M\in \mathbb{R}$ such that $m<\|x_k\|<M$ for all $k$. Now, let $m\le x_{k(1)}:=x_1$, $x_{k(2)}:=\{x_i\in(x_k):\|x_i\|=\max\{\|x_{k(1)}\|$, $\|x_2\|\} \text{ and } 1\le i\le 2\}, \dots$, $x_{k(p)}:=\{x_i\in (x_k):\max\{\|x_{k(p-1)}\|, \|x_p\|\}\}, \dots \le M$. By Bolzano-Weierstrass, $(x_{k(p)})$ converges to some vector in $\mathbb{R}^n$. Suppose that this vector is $a$. Now let $M\ge x_{k(1)}:=x_1$, $x_{k(1)}:=x_1$, $x_{k(2)}:=\{x_i\in(x_k):\|x_i\|=\min\{\|x_{k(1)}\|$, $\|x_2\|\} \text{ and } 1\le i\le 2\}, \dots$, $x_{k(p)}:=\{x_i\in (x_k):\min\{\|x_{k(p-1)}\|, \|x_p\|\}\}, \dots \ge m$. Then this subsequene will converge to a vector $b\in\mathbb{R}^n$, with $b\ne a$ since $\|b\|$ can be taken arbitrarily closely to $m$.
Please let me know what you think about my proof. I'm not very confident that it is quite rigorous. Maybe I misunderstood what actually needs to be proved. Your insight would be much appreciated.
Update: I think the easiest would be to consider subsequences which converge to $\liminf$ and $\limsup$, so there are at least two such subsequences in a bounded sequence.
 A: Hint:  
The sequence must have at least two limit points. Bolzano-Weierstrass guarantees at least one, say $b.$ Since $(x_n)$ is not convergent show there must be an $\epsilon > 0$ and a subsequence $x_{n_j}$ such that $\|x_{n_j} - b \| > \epsilon$ and notice that $(x_{n_j})$ is another bounded sequence.
Addendum
To construct the above subsequence note that the total sequence is not convergent and, in particular, not convergent to $b$. Negating the definition of convergence, there exists $\epsilon > 0$ such that for every $N \in \mathbb{N}$ there exists an integer $n \geqslant N$ such that $\|x_n - b \| > \epsilon$.
With $N = 1$ there exists $n_1 \geqslant 1$ such that $\|x_{n_1} - b\| > \epsilon$. Using induction, suppose we have $n_1 < n_2 < \ldots < n_m$ such that $\|x_{n_j} - b \| > \epsilon$ for $1 \leqslant j \leqslant m$. There must exist an integer $n_{m+1} \geqslant n_m +1 > n_m$ such that $\|x_{n_{m+1}} - b \| > \epsilon$. By induction, we have constructed a subsequence $(x_{n_j})$ that fails to converge to $b$.  This subsequence is bounded and thus has a convergent subsequence (which also is a subsequence of $(x_n)$) that converges to some $b' \neq b$. Hence, there are at least two limit points.
