 Solution

Let $d(u)$ be the weight of a shortest $vu$-path, and let $p(u)$ be the predecessor of $u$ on the chosen $P_u$. Its prefix to $p(u)$ must also be shortest: a cheaper prefix followed by the final edge would give a cheaper walk to $u$, and cycle erasure in minimum-weight walks would produce a cheaper path because every edge weight is positive. ThereforeFollowing predecessor arrows strictly decreases $d$, so cannot revisit a vertex. Since each nonroot vertex has a predecessor, finiteness forces every predecessor chain to end at $v$. Thus $T$ is connected.

The edges $e_u$ are distinct: if two vertices selected the same undirected edge, each would be the other's predecessor, contradicting strict decrease. Hence $T$ has exactly $|V(G)|-1$ edges. A finite connected graph with this edge count is a tree. This proves that positive-weight shortest-path predecessors form a tree, without requiring the initially chosen paths to be mutually consistent.