Proof by contradiction:
Let s and t have a minimal path P1 of length |P1| > |v|/2. Assuming that there's no bottleneck node in P1 means that there exists an alternative, disjoint path P2 between s and t (sharing only the nodes s and t). Since P1 has the shortest length, we know that |P2|>=|P1|.
Now, the total number nodes in the graph must be at least the number of nodes in the union of P1 and P2:
|v| >= |P1|-1 + |P2|-1 + 2 = |P1| + |P2| >= 2|P1| > |v|
And this is the contradiction.