A question regarding a $k$-connected graph I would like to show several things, some for general $k$-connected graphs and some for several instances ($k=2,3,...$). 
First, I want to show that for every $k$-connected graph each subset $A\subseteq V$ of size $k$ is on a cycle. 
How do I go on it? I thought to try by induction on the connectivity of the graph ($k$). 
My base case is for $k=2$. I take 2 arbitrary vertices $v_1, v_2$. I know by Menger's theorem that there are two vertex-distinct path $P_1$ and $P_2$, both from $v_1$ to $v_2$. By going through $P_1$ and reversing $P_2$ I get a cycle. So this works for 2 vertices. 
Now I would like to use my induction hypothesis for $k-1$ in order to apply it to a $k$-connected graph. But I'm not sure how. My intuition is to remove a vertex, and look at the subgraph that was left. Then I know that every $k-1$ vertices are on a cycle. Lets look at the vertex I added $v$. Here is where I begin to mumble, because I know $v$ has $k$ paths from it to each vertex on the cycle.. But shouldn't 2 paths be enough? Why do I need all the $k$ paths? And where does it come to use that they are vertex-distinct?
I was trying to even demonstrate it (the induction step) on a $3$-connected graph. But it just seems its enough to have 2 paths and not 3... 
Another interesting variant is what happens if I take a subset of not only vertices, but also edges. This means I have a subset of size $k$ of vertices and edges. Specifically, I tried focusing on a basic case of $k=3$, and see if I can extend it from there.
So for $k=3$ I want to prove every subset of vertices and edges is on a cycle. I have understood that if the subset is only edges, this might not hold, so I assume it includes at least one vertex. My idea is to look at the edge not as an edge, but rather on its two endpoints and then try to show its endpoints, along with the other vertices in the subset, are on a cycle. But wouldn't it be like showing 4 vertices are on a cycle in a $3$-connected graph? I want to somehow make use of the edge between the two of them, but I am not sure exactly how.
 A: Here is a more detailed argument for the induction step (it is the same argument as here).
Let $A$ be of size $k$. Choose $x\in A$. We know $G$ is also $(k-1)$-connected and thus there exists a cycle $C$ in $G$ containing $A\setminus\{x\}$. If $x$ is in the cycle $C$, we are done. Now suppose $x\notin C$.
If you read here, you will see a stronger form of Menger's theorem. More precisely, if $A,B$ are sets of vertices of $G$, then

In other words, if no k−1 vertices disconnect A from B, then there exist k disjoint paths from A to B.

Thus, we can choose $k$ disjoint paths between $N(x)$ and our cycle $C$. From these, we can find $k$ paths between $x$ and $C$ which are disjoint (except for the vertex $x$), by either appending $x$ to the start of the path or, if $x$ is already in a path, removing all vertices appearing before it in the path. Let us name these paths $P_1,\dots,P_k$ and denote $v_1,\dots,v_k$ respectively the last vertices of the respective paths, these are all disjoint vertices.
Now, as there are $k-1$ elements of $A\setminus\{x\}$ in $C$, there must exist at least two vertices $v_i,v_j$ such that, in the cycle $C$, no vertex of $A\setminus\{x\}$ appears "between" $v_i$ and $v_j$. This means that the part of the cycle between $v_i$ and $v_j$ does not contain any vertex of $A\setminus\{x\}$. Now form $C'$ by appending the part of the cycle that excludes the vertices between $v_i$ and $v_j$, then the path from $v_j$ to $x$, then the path from $x$ to $v_i$.
By construction, the part of $C$ that we did not delete contained all vertices of $A\setminus\{x\}$ and we added $x$, thus $C'$ contains $A$. Furthermore, $P_i$ and $P_j$ are distinct and do not contain any vertex of $C'$ other than $v_i$ and $v_j$, thus $C'$ is indeed a cycle.
