Let $G$ be a $k$-regular graph, with $|V(G)|$ even, that remains connected when any $k − 2$ edges are deleted. Prove that $G$ has a $1$-factor.
I'm not looking for someone to give me a formal proof, I'm just looking for ideas as to how I could go about proving this, I can then fill in the details myself.
In general, Tutte's theorem is the way to guarantee a perfect matching (or $1$-factor) in an arbitrary graph. Given any set of vertices $U$, you want to show that $G - U$ has at most $|U|$ odd components.
The key is that whenever $C$ is a set of vertices that form an odd component of $G-U$, you have a cut that you can apply your hypothesis to: the edges between $C$ and $U$. There cannot be too few such edges (because $G$ is $(k-1)$-edge-connected). However, there also cannot be many odd components, each with many edges incident to $U$: there are only so many edges out of $U$ to go around.
Play these off against each other - making sure to use the parity of $|C|$ at some point - and you'll satisfy the hypotheses of Tutte's theorem.
(Once you do, I encourage you to post your solution as an answer to your own question.)