Prove $x_n$ converges IFF $x_n$ is bounded and has at most one limit point I'm not entirely sure how to go about proving this so hopefully someone can point me in the right direction. The definition I have for a limit point is "$a$ will be a limit point if for a sequence $x_n$ there exists a subsequence $(x_{n_k})$ such that $\lim_{k\to\infty} (x_{n_k})=a$". 
In the forward direction if $x_n$ converges to a value $a$ then every convergent subsequence of $x_n$ must also converge to $a$. I'm not entirely sure where else to go from here. Intuitively I know that what I'm trying to prove must be true since if $x_n$ converges to $x$ then there exists an $N\in\mathbb{N}$ such that $\forall n\geq N$ we have infinitely many terms satisfying $\forall \varepsilon >0\Rightarrow x-\varepsilon<x_n$ which implies that $x$ is a limit point. I'm not entirely sure how to rule out a second limit point. Would it be sufficient to show that if $a$ and $b$ are limits of a convergent sequence $x_n$ then $a$ must equal $b$?
In the backwards direction if $x_n$ is bounded and has only one limit point then we know that there exists only one point $x$ such that we can find infinitely many terms satisfying $\forall \varepsilon >0$ $\left | x_n-x\right |<\varepsilon$ which implies that $x_n$ converges to $x$. 
I'm quite concerned I am proving this incorrectly since I don't think I am using any facts about subsequences although they are mentioned directly in the definition for a limit point.

First show: a point $x_0$ is a limit point of a sequence $\{x_n\}$ if and only if for every $\varepsilon>0$, there exists an open ball $B(x_0,\varepsilon)$ centered at $x_0$ with radius of $\varepsilon$ that contains infinite number of $x_n$.
In the leftward direction of the original problem, let $x_0$ be the unique limit point. For every $\varepsilon>0$, there exists an open ball $B(x_0,\varepsilon)$ that contains infinite many of $x_n$. Thus, there are finite many $x_n$ that are outside the open ball. Otherwise, these infinite exceptional points have another limit point, which is not the case since $x_0$ is the unique limit point. Let $N$ be the max index of these exceptional points. Therefore, for all $n>N$, we have $x_n\in B(x_0,\varepsilon)$, which proves $x_n\rightarrow x_0$ as $n\rightarrow \infty$. 
