Show that if G is a simple graph with at least 4 vertices and 2n-3 edges, it must have two cycles of the same length.

Jernej is correct, $|V(G)| = n$ and $|E(G)|=2n-3$ does not imply that $G$ is connected. Counter example: Consider $n = 6$, then $2n-3=9$. You can construct a graph consisting of a single isolated node and an almost completely connected component with 5 vertices. This is possible because $|E(K_5)| = 10 > 9$.

WLOG let $G$ have $k$ connected components $c_i$, $i=1,\dots,k$. Let $c_j$ be such that $|V(c_j)| \geq 4$ and $|E(c_j)| \geq 2\cdot|V(c_j)| - 3$. Such a $c_j$ must exist because $G$ could not have $2n-3$ edges otherwise. Now consider $T$, the spanning tree of $c_j$. It consists of $|V(c_j)| - 1$ edges, so there are at least $|V(c_j)| - 2$ edges left. Each of these edges creates a cycle of length between 3 and $|V(c_j)|$ when joined with $T$. Since there are $|V(c_j)| - 2$ edges and $|V(c_j)| - 2$ possible distinct cycle lengths this is not yet enough to apply the pigeon hole principle. However, observe that when there is a cycle of length $|V(c_j)|$ every additional edge will create at least two new cycles. Hence either there will be a repeated cycle length when choosing $|V(c_j)|-2$ lengths with $|V(c_j)|-2$ edges, or if not, then there must be a cycle of length $|V(c_j)|$ and so there will be $2(|V(c_j)|-3)+1$ cycles. $2(|V(c_j)|-3)+1$ > $|V(c_j)|-2$ for all $|V(c_j)| > 3$ so the pigeon hole principle can be applied in this case. $\quad\square$