Proving cluster points are $1$ or $-1$ for $x_n$ if $x_n^2 \rightarrow 1$ as $n \rightarrow \infty$.

Question

Suppose $x_n$ is a sequence and $x_n^2 \rightarrow 1, n \rightarrow \infty$. Suppose $c$ is a cluster point of $x_n$. Prove $c = 1$ or $c = -1$.

Not really sure about how to prove that something is either a or b. Could really use some help with what direction to take this. Also, not sure if using the definition of cluster point correctly in proof. Here is my attempt.

Suppose that $c$ is neither 1 nor -1. Since 1 is not a cluster point, then for $\epsilon = 1 > 0,$ we have that $|x_n - 1| > 1$ for infinitely many $n$. Likewise, since -1 is not a cluster point, for $\epsilon > 0$ there are infinitely many $n$ such that $|x_n + 1| > \epsilon$.

Then we have for infinitely many $n$ that: $|x_n - 1||x_n + 1| > |x_n + 1| > \epsilon.$ Therefore, $|x_n - 1||x_n + 1| = |x_n^2 - 1| > \epsilon.$ However, this is impossible since $x_n^2$ converges to 1 there cannot be infinitely many $n$ such that $|x_n^2 - 1| > \epsilon$. Thus we have reached a contradiction.

Answer 1

Assume $c$ is a cluster point such that $c\ne 1$ and $c\ne -1$. Then there exists a sub-sequence $x_{n_k}\to c$. Then $x_{n_k}^2\to c^2\ne 1$. Contradiction.

EDIT: $\epsilon-$proof

Suppose $c\ne\pm 1$ is a cluster point of $x_n$. Choose $0<\epsilon<\min\{(c-1)^2,(c+1)^2\}$

Then the set $V=[-1-\sqrt\epsilon,-1+\sqrt\epsilon]\cup[1-\sqrt\epsilon,1+\sqrt\epsilon]$ does not contain $c$, so $U=\mathbb R\setminus V$ is an open neighborhood containing $c$. Moreover $$x\in U\iff \left|x-1\right|>\sqrt\epsilon\text{ and }\left|x+1\right|>\sqrt\epsilon$$

Thus, there are an infinite $x_n\in U$ (by cluster property). For these $x_n$, one has $\left|x_n-1\right|>\sqrt\epsilon$ and $\left|x_n+1\right|>\sqrt\epsilon$, and by multiplication one gets $\left|x_n^2-1\right|>\epsilon$. Contradiction with $x_n^2\to 1$