I tried using the contrapositive to prove the original statement:
If no cycle in G is an induced subgraph, then G is a graph with no odd cycles.
To prove this, I assumed that G did have an odd cycle that was not an induced subgraph. However, if the odd cycle is not an induced subgraph, then there is an edge connecting two vertices in the "cycle" that shortens the cycle, in a sense, by bypassing some vertices and edges, which creates a new cycle. If this cycle is even, then there is a contradiction, since G was assumed to have no even cycles. If this cycle is odd, then there is a contradiction, since it is an induced subgraph of G.
Am I thinking about this in the right way?