[Math] If $x,y$ is a 2-edge cut of a graph $G$; every cycle of G that contains $x$ must also contain $y$

Question

$x,y$ 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:

Answer 1

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 $z=y$.