If ${\lvert{E(G)}\rvert}>{\lvert{V(G)}\rvert}+3$, graph $G$ has at least two disjoint cycles. Assume that $G$ is a graph which ${\lvert{E(G)}\rvert}>{\lvert{V(G)}\rvert}+3$. Prove $G$ has two disjoint cycles.

Answer:
I tried to prove it when $G$ is connected. consider the smallest cycle of length $k$. It exists because ${\lvert{E(G)}\rvert}>{\lvert{V(G)}\rvert}-1$.
Now, I divided vertices into $2$ groups. Group $A$ consists of $k$ vertices in the smallest cycle and group $B$ consists of the other $n-k$ vertices.
Now, between  $n-k$ vertices are exactly $n-k-1$ edges. Because, If it is more then we have another cycle (the proof is done) and if it is less then $G$ is disconnected (contradiction). So there are at least $5$ edges between group $A$ and $B$.
Now, what should I do? should I write every possibility to prove it? Are there any easy ways to continue the proof?
What about if $G$ is disconnected? I think we should do it in each component. Am I right?
Thank you so much for your answers.

Suppose the result is false; take a minimal counterexample $G$ with $n$ vertices and exactly $n+4$ edges (if there were more edges, we could delete some).
Then $G$ cannot contain a $3$-cycle or $4$-cycle. Otherwise, removing the edges of such a cycle, we have a graph with $n$ vertices and at least $n$ edges, which must still contain another cycle.
Also, $G$ has minimum degree $3$. If $G$ had a vertex $v$ of degree $1$, we could delete it and the edge out of it, getting a smaller counterexample (since the leaf was not part of any cycle). If $G$ had a vertex $v$ of degree $2$, with neighbors $w_1, w_2$, we could replace $v$ and its two edges with the single edge $w_1w_2$. This also gives a smaller counterexample, since any cycle containing $v$ would have had to use both edges, and now it can use the edge $w_1w_2$ instead. Also, because $G$ had no $3$-cycle, this still gives a simple graph.
Since $G$ has minimum degree $3$ and $n$ vertices, it must have at least $\frac 32n$ edges, but we know it has exactly $n+4$. Therefore $n+4 \ge \frac32n$, or $n \le 8$.
However, if we take a vertex $v$, we know it must have at least $3$ neighbors $w_1, w_2, w_3$. These have no edges between them (because $G$ has no $3$-cycles) and each has $2$ more neighbors (by minimum degree $3$) which are all distinct (because $G$ has no $4$-cycles), giving us at least $10$ vertices already. This is a contradiction, so the result must be true.
