I think I have figured out a proof for the question stated in the title. Was wondering if I could get some opinions/review. Also I feel like there is a simpler proof out there.
Given $\epsilon >0$, define $y_n = \inf\{x_m : m \geq n\}$ then by definition $\lim_{n\rightarrow \infty}y_n = 0$. Thus, there exists an $N$ so that for all $n \geq N$ we have $|y_n| < \epsilon$. Let $N_1 = N$, then $y_{N_1} = \inf\{x_m : m \geq N_1\}$, so $y_{N_1} + \epsilon$ is not a least upper bound. Therefore, there exist some $n_1 \geq N_1$ so that $x_{n_1} < y_{N_1} + \epsilon$. Note that $y_n$ is increasing and converges to $0$ so $y_n \leq 0$. Hence, $x_{n_1} < \epsilon$.
Also, since $y_{N_1}$ is a least upper bound then $-\epsilon < y_{N_1} \leq x_{n_1}$. And so we have $-\epsilon < x_{n_1} < \epsilon$. Now let $N_2 = n_1 + 1$. Repeating the above process, we can construct a subset $x_{n_k} \in (-\epsilon, \epsilon)$ of $x_n$. Thus, $0$ is a cluster point.
Suppose $0$ is not a cluster point. Then there exists $N$ and $\epsilon>0$ such that for $n>N$, $|x_n|>\epsilon$. Therefore $|y_n|\geq \epsilon$ for $n>N$. This contradicts the fact that $y_n\rightarrow 0$.