Can't see how this can be true: any connected undirected graph $G$ contains vertex v such that removing v results in another connected Graph $G'$
Suppose to the contrary that removing any vertex of a finite graph $G$ would disconnect it. Then $G$ could not contain any cycles (else we'd have an immediate contradiction) and so $G$ would have to be a tree. However, removing a leaf vertex of $G$ wouldn't disconnect it, contrary again to our assumption.