G a k-connected graph. Show that a set of k-2 vertices and a set of two edges lie on a common cycle. Let $G$ be a k-connected graph with $k \geq 2$. Let $S$ be a set consisting of two edges and let $W$ be a set of $k-2$ vertices. Prove that there exists a cycle in $G$ that contains all elements of $S$ and $W$. 
I'm trying to set up the proof similarly to the proof that any k vertices in a  k-connected graph lie on a common cycle. I can prove that the two edges lie on a common cycle and then picking the largest cycle which contains the max number of vertices from W. Then I want to show that it must contain every vertex in W. I want to try to assume that it contains less than all k-2 vertices and then show I can add one to get a contradiction. However, I keep running into a bunch of problems when I try to do the portion where I add in the vertices. 
In the case where the two edges are disjoint and their vertices don't lie in W, I'm having issues making sure that it remains a cycle after adding in another vertex from W. 
 A: Let's proceed by mathematical induction.
In the base case, $k = 2$, we have a 2-connected graph and two edges.
If we subdivide each of the two edges with a newly added vertex, the graph remains 2-connected, and the two vertex-disjoint paths between these two vertices combine into a cycle which also exists in the underlying graph.
Now suppose $k > 2$, and that for every $(k-1)$-connected graph there is a cycle through every choice of two edges and $k-3$ vertices.
Let $G$ be a $k$-connected graph, $e$ and $f$ be edges in $G$, and $W$ be a set of $k-2$ vertices.
Let $x \in W$. Since $G$ is also $(k-1)$-connected, there is a cycle $C$ through $e$, $f$, and the vertices of $W \setminus \{x\}$. If $x$ is already on $C$ we are done; suppose otherwise.
Label the vertices in $e$, $f$, and $W \setminus \{x\}$ in the order they appear on $C$ as $v_0, \dotsc, v_k$ such that $v_0$ and $v_k$ are the endpoints of $e$, and let $r$ be the first index of an endpoint of $f$ (so $v_{r+1}$ is the other endpoint of $f$).
Now we partition all the vertices which appear on the cycle $C$ into $k-1$ sets,
by starting with the two arcs between the edges $e$ and $f$, and introducing a split at each element of $W \setminus \{x\}$:


*

*Vertices occurring between $v_0$ and $v_1$ (inclusive)

*Vertices occurring after $v_{i-1}$, up to and including $v_i$, for $i \in \{2, \dotsc, r\}$

*Vertices occurring between $v_{r+1}$ and $v_{r+2}$ (inclusive)

*Vertices occurring after $v_{i-1}$, up to and including $v_i$, for $i \in \{r+3, \dotsc, k\}$.


We can verify that the number of sets is $1 + (r-1) + 1 + (k - (r+2)) = k - 1$.
By Dirac's Fan Lemma, there is a fan from $x$ to the vertices of $C$ of size $k$ (that is, a set of $k$ paths, each from $x$ to a vertex of $C$, such that any two of them share only the vertex $x$). We can cut these paths so that they only touch the cycle once (if they touch it multiple times, just cut the path so it ends at the first vertex in the cycle).
By the pigeonhole principle, two of the paths in the fan enter $C$ in the same one of the $k-1$ sets we split it into. Let $u, u'$ be the vertices where these paths reach $C$. Replacing the $u, u'$-portion of $C$ by the $x,u$-path and the $x,u'$-path in the fan builds a new cycle that contains $x$, all of $W \setminus \{x\}$, and the edges $e$ and $f$.
Proof modeled on, and occasionally quoted from, the proof of Dirac's Theorem given by D.B. West in Introduction to Graph Theory, 4.2.24
