Here is a proof by (strong) induction which does not use the fact that a graph with minimum degree at least $2$ has a cycle.
Let $T(n)$ denote the statement: every graph with $n$ vertices and at least $n$ edges has a cycle. Assuming that $T(m)$ holds for all $m\lt n,$ we prove that $T(n)$ holds.
Let $G$ be a graph with $n$ vertices and at least $n$ edges. Choose a vertex $v$ of $G.$ Let $G_1,\dots,G_k$ be the connected components of $G-v,$ and let $G_i$ have $n_i$ vertices and $e_i$ edges.
Case 1. For some $i$ we have $e_i\ge n_i.$
Then $G_i$ has a cycle by the inductive hypothesis $T(n_i).$
Case 2. For some $i$ there are at least two edges from $v$ to $G_i.$
So there are two edges joining $v$ to vertices $x,y$ in $G_i.$ Since $G_i$ is connected there is a path connecting $x$ to $y$ in $G_i,$ which together with the two given edges makes a cycle.
Case 3. For each $i$ we have $e_i\le n_i-1,$ and there is at most one edge from $v$ to $G_i.$ Then the degree of $v$ is at most $k,$ and the total number of edges in $G$ is at most
contradicting our assumption that $G$ has at least $n$ edges.