I'm studying topology by Munkres.
Theorem 27.1:
Let $X$ be a simply ordered set having the least upper bound property. In the order topology, each closed interval in $X$ is compact.
What makes me confused is that if I consider $X = (-\infty,0]$, it has the upper bound property for sure, and $X$ is a closed interval, because it's the complement of empty, which is open in any topology.
So, if I consider the collection $\mathcal{A} = \{(-n,0] \mid n \in \mathbb{N}\setminus \{0\}\}$ as a covering of $X$, I can't find any finite subcollection of $\mathcal{A}$ that covers $X$. I'm supposing that $(-n,0]$ are open in $X$, or shouldn't I?
Thank you in advance for any help.