The first half of this is easy: \(T\) is connected, because there is a path between every pair of vertices. To show that \(T\) has no cycles, we assume it does, for the sake of contradiction. Let \(u\) and \(v\) be two distinct vertices in a cycle of \(T\text{.}\) Since we can get from \(u\) to \(v\) by going clockwise or counterclockwise around the cycle, there are two paths from \(u\) and \(v\text{,}\) contradicting our assumption.
in-context