Construct a sequence $(a_n)_{n=1}^{\infty}$ which has exactly three limit points, at $- \infty, 0,$ and $+ \infty$

Definition. Let us say that a sequence $(a_n)_{n=M}^{\infty}$ of real numbers has $+\infty$ as a limit point iff it has no finite upper bound, and that it has $-\infty$ as a limit point iff it has no finite lower bound.

Exercise. Using the above definition, construct a sequence $(a_n)_{n=1}^{\infty}$ which has exactly three limit points, at $- \infty, 0,$ and $+ \infty$.

The above exercise comes from Analysis I, 3rd edition, Terence Tao. I have presented my solution below.

For Part I below, I construct the sequence and try to rigorously show that it has the said limit points (I really only show that $+\infty$ and $0$ are limit points. To show $-\infty$ would follow a similar proof for $+\infty$). For this part, I am hoping someone can make sure my proofs are valid.

For Part II below, I tried to show that $- \infty, 0,$ and $+ \infty$ are the only limit points. Here I had trouble. Could someone let me know if there's any way I can show this based on my constructed sequence and based on my attempt? Any hints?

Part I

Let $D_1 = \{1, 2, 4, 5, 7, 8, \ldots\}$ and let $D_2 = \{3, 6, 9, 12, \ldots\}$. Consider the sequence $(a_n)_{n=1}^{\infty}$ defined by

$$a_n = \begin{cases}n(-1)^n & \ \text{if} \ \ n \in D_1 \\ \tag{1}0 & \ \text{if} \ \ n \in D_2\end{cases}$$

Thus, the sequence is

$$-1 ,2, 0, 4, -5, 0 ,-7,8 ,0, \ldots$$

I'm going to show $+ \infty$ is a limit point by showing the sequence has no finite upper bound.

Suppose, for sake of contradiction, the sequence has a finite upper bound $M \in \mathbb{R}$, such that $a_n \leq M$ for all $n \geq 1$. There's a result which states that there exists a positive integer $N$ such that $M \leq N$. For this $N$, there must exist an even $N' \in D_1$ such that $N < N'$, implying $M \leq N < N'$. Observe that, since $N'$ is even,

$$N' = N'(-1)^{N'} = a_{N'}$$

where the second equality follows from $(1)$. Since $M \leq N < N' = a_{N'}$, we have that $M < a_{N'}$. But this contradicts that $a_n \leq M$ for all $n \geq 1$. Thus, we must have that the sequence has no finite upper bound. By the definition above, $+ \infty$ is a limit point of the sequence. By an analogous argument, $- \infty$ is also a limit point of the sequence.

Now I show that $0$ is a limit point of the sequence. We have to show that for every $\varepsilon > 0$ and every $N \geq 1$, there exists an $n \geq N$ such that $|n(-1)^n - 0| = |n(-1)^n| \leq \varepsilon$.

Let $\varepsilon > 0$ and let $N \geq 1$. Choose an $n \in D_2$ such that $n \geq N$. By $(1)$, this choice of $n$ implies $|n(-1)^n| = |0| = 0 \leq \varepsilon$. Thus, $0$ is a limit point of the sequence too.

Part II

Now I try to show $- \infty, 0,$ and $+ \infty$ are the only limit points of the sequence. Suppose, for sake of contradiction, $c \in \mathbb{R}$ is a limit point where $c \neq 0$. This implies that for every $\varepsilon > 0$ and every $N \geq 1$, there exists an $n \geq N$ such that, by $(1)$, either

$$|n(-1)^n - c| \leq \varepsilon, \ \ \text{if} \ n \in D_1 \tag{2}$$

or

$$|0 - c| \leq \varepsilon, \ \ \text{if} \ n \in D_2 \tag{3}$$

If it's case $(3)$, then it implies that $c=0$. But this contradicts that $c \neq 0$.

Now consider case $(2)$. It implies that $c - \varepsilon \leq n(-1)^n \leq c + \varepsilon$. To show a contradiction, I tried to show there exists an $N \geq 1$ such that $|n(-1)^n - c| > \varepsilon$ for every $n \geq N$. I was successful in showing this only for even $n \in D_1$, so that does not work.

And now I am stuck.