It's actually not that hard. The proof goes like this:
1) Removing all the vertices on the shortest path will render s-t disconnected (obviously, since there is only $< V/2+1$ vertices left, which means any path is $< V/2$ edges, if it were a s-t path, it would contradict the shortest s-t path)
2) We construct an oriented graph, where
V' = \{s, t\} \cup \{ v_+, v_- | v \neq s, t\}
E' = \{(x_-, y_+) | xy \in E\} \cup \{(x_+, x_-)| \{x_+, x_-\} \subset V' \} \cup \{(s, x_+)|sx \in E\} \cup \{(x_-, t) | xt \in E\}
I assume st is not an edge in the original graph; if it were, it could only have less than 2 vertices, which contradicts there are vertices $s$ and $t$).
it can be easily shown any s-t path in the original graph corresponds to a s-t path in the new directed graph and vice versa.
3) max flow = min cut
We have a s-t flow in the new graph (flowing along the shortest path; capacities are all set to 1). It is maximal, because it disallows any other flow through its verticies (bc. $x_+\rightarrow x_-$ capacity is 1) and we know we cannot get from s to t without going through one of the vertices of the shortest path.
Max flow has size 1, therefore, the min cut has also size 1.
4) We use the min cut - either it will correspond to an edge in the original graph (OK, bomb out any of its vertices provided it is not s or t) or it will correspond to a vertex (obvious).