k-connected graph, there exists a cycle that contains any 2 edges and any $k-2$ vertices

Question: Let $$G$$ be a $$k$$-vertex-connected graph with $$k\geq2$$. Let $$S$$ be a set of two edges and $$W$$ a set of $$k-2$$ vertices. Prove that there exists a cycle in $$G$$ containing elements of $$S$$ and $$W$$.

My setup: Label the edges of interest: $$S=\{e_1,e_2\}$$. I know that in a $$k$$-connected graph, any set of $$k$$ vertices must be contained in a common cycle, so we can construct a cycle $$C$$ using the $$k-2$$ vertices of $$W$$ and the last 2 (guaranteed) vertices I can choose to be an endpoint of $$e_1$$ and an endpoint of $$e_2$$. We label the vertices of $$C=\{v_1,v_2,...,v_\ell\}$$ in "cycle order" ($$v_i$$ is adjacent to $$v_{i+1}$$ in $$C$$), where $$\ell\geq k$$.

What I want to do is guarantee that we can include the other endpoint of $$e_1$$ and the other one of $$e_2$$ if they're not already included, maybe by using the Fan Lemma to guarantee that the path to that endpoint is disjoint from the rest of $$C$$ (but I need it to be disjoint from more than $$k$$ points so I'm not sure if this would work). I also think I'd have to introduce a bunch of different cases: the other endpoint of $$e_1$$ is already in $$C$$ and/or the other endopint of $$e_2$$ is already in $$C$$. Not only that, but I'm not sure how to guarantee that even if the endpoints are already in $$C$$, that we can "rearrange" the cycle so that those endopints are adjacent (in $$C$$) via $$e_1$$ or $$e_2$$. It does seem a little unreasonable to introduce all of these cases so I'm looking for a more concise way to approach the problem.

The argument is essentially the same as for the proof that Every $k$ vertices in an $k$ - connected graph are contained in a cycle.
First argue that there's a cycle containing $$e_1$$ and $$e_2$$. (Hint: if you add a vertex $$v_1$$ adjacent to both endpoints of $$e_1$$ and a vertex $$v_2$$ adjacent to both endpoints of $$e_2$$, the resulting graph is still $$2$$-connected.) Then follow the argument in the answers to the linked question to insert the vertices in $$W$$, one at a time.
• Thanks for the hint, I was able to easily show that $e_1$ and $e_2$ are in a common cycle. However, what I am confused by is being able to add all $k-2$ vertices of $W$. If we start the cycle with the at least $3$ vertices of C, how can we be guaranteed to be able to construct a cycle containing at least $k+1$ vertices (e.g. how can I be sure I can add the last vertex, or 2 vertices if $e_1$ and $e_2$ don't share an endpoint) to the cycle? Mar 29 '20 at 4:55