Maybe I've come up with something.. The sum of deg(v) has to be 8 (I have 4 edges, each is shared by two vertices). Moreover, there are 5 vertices, so the only combinations of degrees are: (1,1,1,1,4), (1,1,2,2,2) and (1,1,1,2,3).
(i) There's no other way how to connect (1,1,1,1,4) vertices to yield a nonisomorphic graph with the (1,1,1,1,4) graph from point (a). This is quite clear.
(ii) The same for a (1,1,2,2,2) graph.. This is a path and there's no other nonisomorphic connected graph that could be constructed from these vertices (eg. a graph consisting of a triangle and a path of length 1 has also vertices of degrees (1,1,2,2,2)..but it isn't connected).
(iii) To ensure that a (1,1,1,2,3) graph is connected, the vertices with degrees 3 and 2 must share an edge (otherwise, there would be no path from the former to the latter.. since the rest of the vertices are of deg=1.. and that would yield a graph that's not connected).. thus all graphs with the degree set (1,1,1,2,3) are isomorphic with the one on the picture.
Do you think this is sufficient?