K critical graphs connectivity and cut vertex Show that a k-critical graph is connected. Furthermore, show that it does not have a vertex whose removal disconnects the graph (such a vertex is known as a cut vertex).
I have managed to proove , I think, the first part 
 Let's assume G is not connected. Since χ(G) = k, (If G1,G2,...,Gr are the components of a disconnected graph G, then χ(G) = max χ(Gi) 1≤i≤r ) then there is a component G1 of G such that  χ(G1)=k.If v is any vertex of G which is not in G1,then G1 isa component of the subgraph G − v. Therefore, χ (G − v) = χ (G1 ) = k. This contradicts the fact that G is k-critical. Hence G is connected.
But I can't manage to prove the second part..ie that there is not cut vertex. 
Anyone can help?

Recall that "critical" also means that you cannot remove an edge without reducing the chromatic number of the graph.
If G had a cut vertex you could remove an edge adjacent to that vertex without decreasing $\chi(G)$.
I hope this suffices as a hint.
