G is k-connected, then there is a cycle in G containing all $x_i$. Let $G$ be a $k$-connected graph $(k ≥ 2)$, and let $x_1,x_2,...,x_k$ be vertices of $G$. Show that there is a cycle in $G$ containing all the $x_i$.
First, Assum that $G$ is $k$-connected graph $(k ≥ 2)$ on $n$ vertices.
So  $G-x_1-x_2...-x_k$ is disconnected then we have only two components.
How i prove that there is a cycle in $G$ containing all the $x_i$ for all $i$.
Maybe suppose that contradict? 
 A: One way to do this is by induction on $k$. To go from $k-1$ to $k$, suppose that $C$ is the cycle containing $v_1, \dots, v_{k-1}$ and $v_k \notin C$. We'll use Menger's theorem to make the desired cycle from $C$:


*

*If $|C| \geq k$, there are $k$ internally disjoint paths from $v_k$ to $C$, each ending on a different endpoint in $C$. At least two consecutive endpoints (say the endpoints of paths $p_i$ and $p_j$) do not have any of $v_i$ $(1 \leq i \leq k-1)$ on the path between them in $C$ (why?), and so one can replace it with $p_i$ and $p_j$ to get a cycle that contains all $v_1, \dots, v_k$.

*If $|C| = k-1$, there are $k-1$ internally disjoint paths from $v_k$ to $C$, each ending on a different $v_i$. Take any two adjacent endpoints, and replace the edge between them in $C$ with their paths to $v_k$.
It remains to prove for $k=2$. This case is well-known (actually a stronger version holds: $2$-connected $\iff$ every two vertices are in a cycle). That being said, we could still use Menger's theorem to say there are two internally disjoint paths from $v_1$ to $v_2$, the union of which makes a cycle.
