My question stems from the article "M. D. Plummer, On the cyclic connectivity of planar graphs" (1972). Before asking my question, the following necessary definitions and explanations are provided.
Let $G$ be a finite simple graph.
A set of edges $L$ in $G$ is a cyclic cutset of $G$ if $G-L$ has two components each of which contains a cycle.
The minimum cardinality taken over all cyclic cutsets of $G$ is called the cyclic connectivity of $G$ and is denoted $\mathrm{c} \lambda(\mathrm{G})$.
If no such $\mathrm{L}$ exists in $G$, we set $\mathrm{c \lambda}(\mathrm{G})=\infty$.
The author asserts, without proof, the following:
Assert 1. Let $G$ be a 3-connected simple graph. Then $\mathrm{c} \lambda(G)=\infty$ if and only if $G$ contains no two vertex-disjoint cycles.
I think if-part is easily verifiable. Suppose $\mathrm{c} \lambda(G)<\infty$， i.e. $G$ has a cut set $S$ such that $G-S$ has two components each of which contains a cycle, namely $C$ and $C'$. Obviously, $C$ and $C'$ are vertex-disjoint cycles in $G$. Contradiction.
So we only need to prove that only-if part：If $\mathrm{c} \lambda(G)=\infty$, then $G$ contains no two vertex-disjoint cycles. I don't know how to prove it. But I know the condition of being 3-connected may be necessary through the following example, since the shown graph has $\mathrm{c} \lambda =\infty$, but has two vertex-disjoint triangles.
But I don't know how to incorporate 3-connectivity into the proof, or whether changing the condition to minimum degree 3 would suffice
Edit: As NaturalLogZ and Thomas Lesgourgues remind, the question is easy. In fact, I wrongly thought Plummer considered the cuts being about vertex cuts. My question is based vertex-version.
Does cyclic connectivity have vertex-version? Do we have a similar assert for vertex cyclic connectivity?
Letting $c\kappa(G)$ denote the cyclic-vertex connectivity of a graph $G$ the statement, "$c\kappa(G)=\infty$ if and only if $G$ contains no two vertex-disjoint cycles" is incorrect. Your example can be modified by connecting each node in on of the triangles to its counterpart in the other (and thus obtaining the skeleton of a triangular prism). This is 3-connected, and for the reasons you have suspected, it should have infinite $c\kappa$.
However, the problem is analogous to the problem that arises when talking about the size of a separating set for adjacent vertices. The difference is you can still easily define the connectivity of the whole graph (as the minimum size of a separating set) so long as the graph isn't complete. And since that is just one family of graphs where talking about vertex connectivity is difficult, you can just define the connectivity of a complete graph to equal the number of vertices.
Thus I would argue that the correct translation of assertion 1 into vertex-cyclic-connectivity would be "$c\kappa(G)=\infty$ if and only if $G$ contains no two non-adjacent vertex-disjoint cycles." (Here two subgraphs $H_1,H_2$ are said to be adjacent if there exists $u \in H_1$ and $v \in H_2$ such that $u$ and $v$ are adjacent.) The proof of this statement is identical to the proof for edge-cyclic connectivity, where the one direction was given by you and the other by @ThomasLesgourgues.