If $x_n$ is a bounded sequence of distinct real numbers such that its range has exactly one cluster point, $x_n$ is convergent. I first tried proving this by assuming the cluster point was the limit of the sequence and then assumed that $x_n$ had more than one cluster point, leading to a contradiction. This attempt was totally incorrect.
 A: We say that $x_n$ clusters a point $p$ if each neighborhood of radius $r$ centered at $p$ contains infinitely many  points of the sequence $x_n$. Intuitively this is saying that we can always find a point $x_N$ that is less than$\epsilon=r$ away from $p$. Can you see how this relates to the definition of a convergent sequence given that the limit of a sequence is unique?
A: Let $y$ be a cluster point of $(x_n)$, i.e. there is a subsequence $(x_{n_k})_{k\geq 0}$ which converges to $y$. If $(x_n)$ is convergent, it has to converge to $y$. Indeed, any convergent sequence is a Cauchy sequence. Hence $|x_n - x_{n_k}| \to 0$, as $n, k \to \infty$. By triangle inequality, we have $|x_n -y| \leq |x_n - x_{n_k}| + |x_{n_k} - y| \to 0$, as $n, k \to \infty$.
Suppose $(x_n)_{n \geq 1}$ is not convergent. Then $(x_n)_{n\geq 1}$ is not convergent to $y$. We claim that there is a subsequence $(x_{N_j})_{j \geq 1}$ which stays away from $y$, i.e.
$|x_{N_j} - y| > \epsilon$, for all $j \geq 1$. 
Indeed, since $\lim x_n \neq y$, there is an $\epsilon > 0$ such that for all $N \in \mathbb{N}$, there is a $j > N$ such that $|x_{N_j} - y| \geq \epsilon$.  
We simply extract this subsequence $(x_{N_j})_{j \geq 1}$ from the above observation. This new sequence is bounded because it is a subsequence of a bounded sequence. By property of $\mathbb{R}$, any bounded sequence has a convergent subsequence. Extract the convergent subsequence $(x'_{N_j})$ from $(x_{N_j})$ and $\lim x'_{N_j} = w \neq y$. 
$w$ cannot be $y$ because if so, $x'_{N_j}$ must be arbitrarily close to $y$, which is not the case since $|x'_{N_j}-y| > \epsilon$ for all $j$.
So we get a new cluster point $w$ different from $y$. 
A: Let $\left( x_n \right)_{n\in \mathbb{N}}$ be a bounded sequence of real numbers having distinct terms such that the range $\left\{ x_n \colon n \in \mathbb{N} \right\}$ has exactly one limit point, say $b \in \mathbb{R}$. Then we can show that any two distinct convergent subsequences of $\left( x_n \right)_{n \in \mathbb{N} }$ converge to $b$, because the limit of any convergent subsequence will automatically be a limit point of the range of $\left( x_n \right)_{n \in \mathbb{N} }$, and $b$ is the only limit point of this range.
If every subsequence converges to $b$, then we are done.
On the other hand, if some subsequence does not converge to $b$, then that subsequence has a subsequence---and this latter sequence of course is a subsequence of our original sequence---all of whose terms remain at a certain minimum $\epsilon_0 > 0$ distance from $b$, but this last sequence (i.e. that particular subsequence of the subsequence not converging to $b$) is also bounded and has distinct terms, which implies that its range being a bounded infinite subset of $\mathbb{R}$ has a limit point, and this limit point is also the limit point of the range of our original sequence and so must equal $b$, a contradiction.
Hence every subsequence of $\left( a_n \right)_{n \in \mathbb{N} }$ converges to $b$, as required.
