If $x,y$ is a 2-edge cut of a graph $G$; every cycle of G that contains $x$ must also contain $y$ $x,y$ are cut edges; if I understand the definition, it means that if we delete both $x$ and $y$ our resulting graph is disconnected.
I'm very confused because I started like this:
Let $C$ be a cycle that contains $x$ but not $y$...(here I'm thinking how to get a contradiction with the fact that they are cut edges)...I'll make a couple of drawings...

Oh-oh, this one doesn't work! If I remove both, the graphs ends up disconnected, yet they don't belong to the same cycle.
I don't know if I'm having logic problems here or the definition of vertex cuts needs the cut to be minimum, because in this case just removing $y$ would be enough!
Definitions used in the book:


Let $x=uv, u\in S$, $v\in \overline S$ and $y=wz, w\in S$, $z\in \overline S$. Removal of $x,y$ make the graph disconnected. Suppose there exists a cycle  $\mathscr C$ consists of $x$ without $y$. So, there exists a $u-v$ path other than edge $x$. Let it be $u e'_1u_1 u_2e'_2 ...u_k e'_k v$. There exist $u_i\in S$ and $u_{i+1}\in \overline S$. So, $u_{i}u_{i+1}$ is an edge cut by definition other than $y$. which is absurd. Removal of $x,y$ alone makes graph disconnected.
