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:

 A: You missunderstood the definition of a cut. I hope is clear now thanks to @Gregory J. Puleo.
That being said here is my answer:
Let $C$ be a cycle that contains $x$ and that does not contain $y$. Since $x\in [S,\overline{S}]$ and $x \in C$  it must exist another edge $z\in[S,\overline{S}]$ to complete the cycle between both "partitions". So it must be the case that $z=y$.
A: 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.
A: @N.Maneesh i tried following your argument and well the choice of vertex edge vertex vertex edge notation is confusing to say the least. 
Heres an argument of what the poster was actually trying to prove. (which is hard to tell from the confused diagram he posted that is an incorrect example of his question; though i believe you understood that.)
Question: Prove that If x,y is a 2-edge cut of a graph G; then every cycle of G that contains x must also contain y.
Consider $G'= G \backslash y $  where the edge $x= uv$ then G' is by definition a graph with a one edge cut namely edge x. Your question amounts to is there a cycle starting at u or v in the graph $G'?$. 
Since x is an edge cut there is no path from u to v except x so $G'$ contains no cycle starting at u or v, giving the result.
I know that your question @N.Maneesh was does my solution work? 
i don't like it aside from choice of notation the idea is good until here "$u_{i}u_{i+1}$ is an edge cut is an edge cut by definition other than $y$. which is absurd." i mean i am being a little technical here but this isn't a full marks answer you should say consider G remove x,y then we still have path "insert path you built " then say contradiction of the assumption G is 2 edge cutable; cause you have shown a connection. 
Anyway you should add something about why its absurd the fact that you found a cut edge is the important bit buts its not clear you understand why its important.
