 Solution

We first prove the required case of Menger's theorem directly. Split each vertex $v$ into $v_{min}$ and $v_{mout}$ joined by a directed edge of capacity one. Replace every graph edge $uv$ by the two directed edges $u_{mout}{v}_{min}$ and $v_{mout}{u}_{min}$ of capacity $k$, add a source joined to every $a_{\mathrm{in}}$ for $a\in A$, and join every $b_{\mathrm{out}}$ for $b\in B$ to a sink, using unit capacities at the ends.

Starting with zero integral flow, repeatedly augment by one along a source-to-sink path in the residual directed graph. If fewer than $k$ augmentations have occurred and no augmenting path remains, let $R$ be the residual-reachable vertices. The capacity from $R$ to its complement equals the current flow: every forward edge in this cut is saturated and every reverse edge carries zero flow, by residual reachability. Since the cut has capacity below $k$, it uses no capacity-$k$ graph edge. The original vertices whose unit-capacity split or endpoint edges cross the cut form an $AB$-separator of size below $k$, contrary to the hypothesis. Hence at least $k$ augmentations occur. The resulting integral flow decomposes into $k$ source-to-sink paths, and the unit split capacities make their corresponding $AB$ paths vertex-disjoint. This proves

The same augmenting-path proof, with the source vertex given capacity $k$, yields the fan lemma: from a vertex $x$ to any set of at least $k$ vertices in a k-connected graph, there is a $k$-fan with distinct endpoints.

Now let $C$ be a longest cycle in a finite $k$-connected graph $G$. If $C$ is not Hamiltonian, choose $x\notin C$. A $k$-fan from $x$ to $C$ has $k$ distinct attachment vertices. If $|C|<2k$, two consecutive attachment vertices on $C$ are joined by an edge of $C$, because the $k$ cyclic gaps cannot all have length at least two. The two corresponding fan paths form, through $x$, an alternative path of length at least two between those adjacent vertices. Replacing their edge on $C$ by this path creates a longer cycle, a contradiction. ThereforeThis is the Dirac circumference theorem; in particular, a $k$-connected graph contains a cycle of length at least $k$, and when $|G|\ge 2k$ it must contain one of length at least $2k$.

For $k=3$, every 3-connected graph with at least six vertices consequently has a cycle of length at least six. This cannot be improved: the complete bipartite graph $K_{3,m}$ is 3-connected for $m\ge 3$, but every cycle alternates between its two parts and therefore has length at most six. Taking $m$ arbitrarily large defeats every proposed bound above six. Hence