Prove that for all $x\in\Bbb R$, precisely one of the statements $x>0$, $x=0$, or $x

I came across the following exercise while studying Terrence Tao's book Analysis I:

Exercise 5.4.1 Let $x\in\Bbb R$. Show that precisely one of the statements $x>0$, $x=0$, or $x<0$ holds

My Attempt: First let us prove that at least one of these statements must hold. Suppose $x=\operatorname{LIM}_{n\to\infty}a_n$ is a nonzero real number (where $\operatorname{LIM}_{n\to\infty}$ denotes the formal limit of the sequence that follows). Then the sequence $(a_n)_{n\in\Bbb N}$ cannot be equivalent to $(0)_{n\in\Bbb N}$, i.e. $$ \neg(\forall\varepsilon >0 \ \exists N\in\Bbb N \ \forall n\ge N:|a_n|\le\varepsilon)\equiv \exists\varepsilon >0 \ \forall N\in\Bbb N \ \exists n\ge N:|a_n|>\varepsilon.$$The author hints to conclude that $(a_n)_{n\in\Bbb N}$ is either eventually positively or negatively bounded away from zero; namely$$\exists c>0 \ \exists N\in\Bbb N \ \forall n\ge N: |a_n|\ge c \quad \lor \quad \exists c<0 \ \exists N\in\Bbb N \ \forall n\ge N: |a_n|\le c.$$Clearly, there are infinitely many $n\in\Bbb N$ such that $|a_n|>\varepsilon$, but there is no certainty that this holds for all $a_n$. Nor is there any certainty that the claim holds for all $n\ge N$ for some $N$. So how can I conclude? I have no problems showing that only one of these can hold at once.

Thank you for your time, and happy 4^th for those who celebrate :)