Understanding proof that connected graph with $V=E+1$ is acyclic 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.

Let $v$ be  a vertex not belonging to the cycle $c_1c_2\cdots c_k$. Then by connectedness there exists a path the form $v_0v_1\cdots v_r$ with $v_0=v$ and $v_r=c_i$. Among all such paths for $v$, pick one that minimizes $r$. As $v$ is not in the cycle, $r\ge 1$. Associate $v$ with the first edge $vv_1$ of one such shortest path.
If §w§ is another vertex not belonging to the cycle, we likewise associate it with  the first edge $ww_1$ of a shortest path $ww_1\cdots w_s$ for $w$. The claim is that $vv_1$ is not the same edge as $ww_1$. Indeed, for these to be equal, we need $v_1=w$ and $w_1=v$. But then either $v_1\cdots v_r$ is a shorter path for $w$ or $w_1\cdots w_s$ is a shorter path for $v$, contradiction. Hence the above association of vertices with edges is injective.
