Graphs such that $|G| \ge 2$ has at least two vertices which are not its cut-vertices Show that every graph $G$, such that $|G| \ge 2$ has at least two vertices which are not its cut-vertices.

I think Douglas' answer is good except the very first line: 
G doesn't have to be connected for this question to make sense. 
For a disconnected graph G, the argument could be applied to any connected component of G WLOG. If you can prove that a pair of leaves exists that satisfy the condition for a connected graph (as Douglas does), the same must be true for a connected subgraph of a disconnected graph. 
Note also: for the smallest possible $G$ (consisting of 2 vertices and 0 edges), you simply have two isolated vertices. If you remove both of the vertices from such graph, you have an empty graph (not an increased number of connected components). So the theorem still holds.
