Call a choice of one vertex from each pair a transversal. Every transversal is connected, so it has at least $n-1$ edges.

Let $e_{\mathrm{in}}$ count edges within pairs. In a uniformly random transversal, each edge between pairs appears with probability $1/4$; edges within pairs never appear. Thus

$$\mathbb E[\text{edges}]=\frac{4n-4-e_{\mathrm{in}}}{4}\le n-1.$$

Equality is forced. Hence there are no edges within pairs, and every transversal has exactly $n-1$ edges: it is a tree.