Limit point of a set is not the same thing as limit point of a sequence. Look at the actual definitions:
$x$ is a limit point of the set $S$ if every open nbhd of $x$ contains a point of $S\setminus\{x\}$;
$x$ is a limit point of the sequence $\langle x_k:k\in\Bbb N\rangle$ if for every open set nbhd $U$ of $x$, $\{k\in\Bbb N:x_k\in U\}$ is infinite. (I prefer to use the term cluster point: it’s less likely to result in confusion with the very different notion of the limit of the sequence.)
Now consider the sequence $\langle (-1)^k:k\in\Bbb N\rangle$. Let $U$ be an open nbhd of $1$;
$\begin{align*} \{k\in\Bbb N:(-1)^k\in U\}&=\{k\in\Bbb N:(-1)^k=1\}\\ &=\{k\in\Bbb N:k\text{ is even}\}\;. \end{align*}$
The set of even natural numbers is certainly an infinite set, so $1$ is a limit point of the sequence.
If instead we let $U$ be an open nbhd of $-1$, we have
$\begin{align*} \{k\in\Bbb N:(-1)^k\in U\}&=\{k\in\Bbb N:(-1)^k=-1\}\\ &=\{k\in\Bbb N:k\text{ is odd}\}\;. \end{align*}$
The set of odd natural numbers is also infinite, so $-1$ is also a limit point of the sequence.
However, neither $1$ nor $-1$ is a limit point of the set $S=\{-1,1\}$: $(0,2)$ is an open nbhd of $1$ that contains no point of $S\setminus\{1\}$, and $(-2,0)$ is an open nbhd of $-1$ that contains no point of $S\setminus\{-1\}$.
Nor is either $-1$ or $1$ the limit of the sequence: that has yet a different definition.
- $x$ is the limit of the sequence $\langle x_k:k\in\Bbb N\rangle$ if for every open nbhd $U$ of $x$ there is an $n_U\in\Bbb N$ such that $x_k\in U$ whenever $k\ge n_U$.
Neither $-1$ nor $1$ satisfies this definition for the sequence $\langle(-1)^k:k\in\Bbb N\rangle$. Take $U=(0,2)$, for instance; this is an open nbhd of $1$, and no matter how big you set the cutoff $n$, there will be a $k\ge n$ such that $k$ is odd and therefore $(-1)^k=-1\notin U$. A very similar argument shows that $-1$ is not the limit of the sequence.