It suffices to show it’s totally disconnected. This follows from (T45) and (T46), but the whole argument is: Let $x \ne y$. Then $x \in U$ and $y \in V$ with $U \cap V = \emptyset$. So $x \notin \cl{V}$, which is clopen. Then $x$ can’t be in the connected component $C$ containing $y$, since $\cl{V} \cap C$ and $\cl{V}^C \cap C$ would be disjoint nonempty open sets.

By definition.

It’s impossible to have an infinite strictly decreasing sequence of open sets if there are only finitely many possible open sets.

Globally implies locally.