In my textbook, the author claims the following: If $x$ is not a limit point of $(x_{n_k})$ [where $(x_{n_k})$ is subsequence of $(x_n)$] if $$ \exists \epsilon > 0 \; \; \exists N > 0\; \; such \; that \; \; n > N \implies\; \; |x-x_{n_k}| > \epsilon $$
I think this is incorrect and here is my reasoning: Since $x$ is not a limit point, then theres no subsquence of $x_{n_k}$ converging to $x$ so that $x_{n_k}$ itself does not converge to $x$. This means:
$$ \exists \epsilon > 0 \; \; \; such\; that \; \; \forall N > 0 \; \; \exists n > 0 \; \; so \; \; n > N \implies |x_{n_k} - x | \geq \epsilon $$
Now, what is going on here?
Your argument does not say that $x$ is not a limit point of $(x_{n_k})$.
As Masacroso points out in his comment, $x$ is a limit point of a sequence $(\xi_k)$ if and only if for every $ϵ>0$ there are infinitely many $k∈\mathbb N$ such that $|x−\xi_k|<ϵ$.
Take $\xi_k = x_{n_k}$. Then negation leads to:
$x$ is not a limit point of $(x_{n_k})$ if and only if there exists $ϵ>0$ such that there are only finitely many $k∈\mathbb N$ such that $|x−x_{n_k}|<ϵ$. The latter means $$ \exists \epsilon > 0 \; \; \exists K > 0\; \text{such that } \forall k > K : |x-x_{n_k}| \ge \epsilon $$ This is of course equivalent to $$ \exists \epsilon' > 0 \; \; \exists K > 0\; \text{such that } \forall k > K : |x-x_{n_k}| > \epsilon' $$ In fact, given $\epsilon$, take $\epsilon' = \epsilon/2$. Conversely, given $\epsilon'$, then $\epsilon = \epsilon'$ will do.
You correctly state that since $x$ is not a limit point, then there is no subsquence of $x_{n_k}$ converging to $x$ so that $x_{n_k}$ itself does not converge to $x$. However, this is not equivalent to the original statement. It is possible that $x_{n_k}$ does not converge to $x$, but nevertheless has $x$ as a limit point. Therefore your formula is correct, but is not the adequate property.