This is an excerpt proving that if a graph $G$ is connected and has one more vertex than edge, then it is acylic.
Suppose $|V|=n$ and that $G$ has a $k$-cycle. This cycle has $k$ vertices and edges, hence $G$ has $n-k$ additional vertices. Each of these vertices has a minimal path to the cycle. By minimality, each of these paths contains an edge not appearing in any other. Hence we have at least $n-k$ new edges, so at least $n$ in total, contradicting the assumed equality.
Can someone explain how exactly minimality implies each path has an edge not on any other? It seems to me entirely possible that a minimal path is entirely contained in another - simply have one vertex adjacent to the cycle, and another one adjacent to the previous one but not directly to the cycle.
The proof is $3\implies 4$ here.