Suppose for a contradiction that the result doesn't hold.

So, if we remove a vertex, then there is another path from u to v in the graph. So, it follows that these two paths must create a cycle.

If the paths are internally disjoint (all vertices apart from u and v are distinct), then by hypothesis, each path is of length greater than k/2. So, the cycle has more than k vertices, which cannot happen.

So, any two paths between u and v must share at least one internal vertex

this should be enough to get you started.