I would appreciate it if anyone can verify my proof. It is a proof by induction, but I attempt to reason things out rather than using a purely mathematical approach, in a similar vein to many other graph theory proofs I have come across.
Base case: A graph with 1 vertex is defined to be connected. Clearly it has at least 0 edges.
Inductive case: Assume that for some $k \in \mathbb Z^+$, every connected undirected graph with $k$ vertices has at least $k-1$ edges. If we select any of such a graph, and introduce a single vertex $k' $, then there are now $k+1$ vertices. We know that there was a path between every two distinct vertices in the old graph. To maintain connectivity, we must join this new vertex $k'$ to $at$ $least$ $one$ of the $k$ vertices in the old graph. Suppose we join $k'$ to a vertex $v$ in the old graph. Then there will be a path from $k'$ to each of the $k$ vertices in the old graph $via$ $v$. Thus the new graph is connected, and it has $k+1$ vertices and $k$ edges.
Therefore $\forall$ $n\in\mathbb Z^+$, every connected undirected graph with $n$ vertices has at least $n-1$ edges.