$k$-regular even graph, still connected after removing k-2 edges has a 1$-factor I have been asked to show that for a $k$-regular graph with even number of vertices has a $1$-factor if the graph stays connected after $k-2$ edges have been removed.
I do not know how to start this, but I think I could use Tutte's theorem somehow but I'm not sure. Does anyone have any hint or know how to solve it?
I have seen similar posts before but they have never had a complete answer, only hints.
Previous posts:
If a k-regular graph of even order remains connected when any k-2 edges are deleted, it has a 1-factor
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.
 A: Lemma. In a $k$-regular graph $G$, if $S \subseteq V(G)$ with $|S|$ odd, there cannot be exactly $k-1$ edges between $S$ and $V(G)-S$.
Proof. Suppose there are $k-1$ such edges. Then $\sum_{v \in S}\deg(s) = k|S|$ counts each edge between two vertices in $S$ twice, and each edge from $S$ to $V(G)-S$ once, so there are $\frac{k|S|-(k-1)}{2}$ edges between two vertices in $S$. But this simplifies to $k \cdot \frac{|S|-1}{2} + \frac12$, which is not an integer. $\square$

Now we are ready to apply Tutte's theorem. Let $U$ be any nonempty subset of $V(G)$. Then there are at most $k|U|$ edges out of $U$. If $C$ is an odd component of $G-U$, it must receive at least $k$ of those edges: there cannot be $k-2$ or fewer by the connectivity condition, and there cannot be $k-1$ by the lemma. Therefore there can be at most $\frac{k|U|}{k}$ odd components.
If $U = \varnothing$, then the connectivity condition does not apply, because there might just be one odd component $C$ which is the whole graph. This is where we need the assumption that $|V(G)|$ is even.
