In Introduction to Graph Theory, the authors state the following theorem attributed to Gabriel Dirac:
If G is a k-connected graph, $k\ge2$, then every k vertices of G lie on a common cycle of G.
and give the following proof:
Let $S=\left\lbrace v_{1},v_{2},\ldots v_{k}\right\rbrace $ be a set of $k$ vertices of $G$. We show that there exists a cycle in $G$ containing all the vertices of $S$. Among all cycles in $G$, let $C$ be one containing a maximum number $l$ of vertices of $S$. We claim that $l=k$. Assume, to the contrary, that $l < k$. Since $G$ is $k$-connected and $k \ge 2$, it follows that $G$ is 2-connected and so $2 \le l < k$. We may assume that $C$ contains the vertices $v_{1},v_{2},\ldots,v_{l}$ of $S$ and that the vertices of $S$ on $C$ appear in the order $v_{1},v_{2},\ldots,v_{l}$ as we proceed cyclically about $C$. In particular, there is a $v_{1} - v_{2}$ path $P$ on $C$ that contains no internal vertices belonging to $S$. Since $l
I am having trouble with this proof because although I understand that $P_{1}$ and $P_{2}$ do not contain any of the vertices $v_{3},\ldots,v_{l}$, it seems to me that these paths may very well contain other vertices of $C$ in which case it is not at all obvious to me how C' is obtained.
Is the proof lacking or am I simply not getting it?