Every $k$ vertices in an $k$ - connected graph are contained in a cycle.

Here is a rough argument:

Take a cycle $C$ containing as many of the $k$ designated vertices as possible. If cycle has all the k designated vertices then you are done. Otherwise, By Menger's theorem , you can choose k paths from a missing vertex to the cycle $C$. The end points of the paths in $C$ divide $C$ into $k$ segments. One of the segments does not contain any of the designated vertices. Replace that segment by the two paths connecting the ends of the segment to the missing vertex.

It should be an induction statment, the induction is on k.

Base of induction begins with k=2 then we know every $u,v \in V$ have at least two vertex-disjoint paths (this is by Menger's theorem) $P_1=(v,...,u)$ and $P_2=(v,...,u)$ , we notice $P_1P_2$ is a cycle and u and v are both in it which means for every group |S|=k=2 it holds that it is contained in a cycle.

Induction hypothesis is that for k-1 - conectedness the statment holds, which means take any groups of vertices |S|=k-1 you'll have them sitting on a cycle C. So now it's left to take arbitary $v \in V$ in a k-connected graph remove it and this gives us a graph k-1 connected so a group |S|=k-1 sits on a circle and now it's only left to look at v-as suggested above using menger and creating a cycle.