Disconnectedness of closed intervals in Sorgenfrey's line

Question

In order to prove Sorgenfrey's line is totally disconnected I took the long road and proved every type of subset except singletones (intervals and rays) is disconnected. Everyone except for closed intervals were easy. I can't seem to prove that a closed interval $[a,b]$, for $a<b \in \Bbb{R}$, is disconnected. That is, I can't find two non-empty open sets (again - under the lower limit topology) whose disjoint union is equal to $[a,b]$.

Any hints?

Answer 1

Since $[a,b]=[a,b)\cup\{b\}$ and $[a,b)$ and $\{b\}$ are open disjoint subsets of $[a,b]$, you're done. Note that $\{b\}$ is an open subset of $[a,b]$ because $\{b\}=[a,b]\cap[b,b+1)$.

Answer 2

All sets of the form $[a,b)$ are clopen (open and closed at the same time) and as these form a base for the topology, $\Bbb S$ is zero-dimensional topologically and Hausdorff, which implies that it is totally disconnected: no subset larger than a singleton is connected.