The sphere with centre $A$ and radius $3$ intersects the three edges at $B$ in three points $K, L, N$. By straightforward calculation using Pythagoras' Theorem, it's easy to show that they are the midpoints of the edges. Let $M$ be an interior point of any of the arcs $KL, LN, KN$ arising from the intersection with the sphere. The sphere with centre $M$ and radius $3$ contains the point $A$. All the other points of the cube are in the interior of this sphere since the distance from $M$ to all the other vertices are $< 3$. Thus $A_2$ must be one of $K, L, N$.

From each of $K, L, N$ we can reach one of the vertices adjacent to $A$. Thus $A_3$ is a vertex connected to $A$ by a single edge. Now $A_4$ must be a midpoint of a side and $A_5$ is a vertex connected to $A$ by at most $2$ edges. Since $A$ is connected to $B$ by a sequence of $3$ edges, we see that $k \ge 7$.

It is easy to construct a sequence of length $7$ that works.