Your proof would be absolutely fine if we can prove that the construction of a simple edge disjoint cycle is always possible. Take any vertex u with a positive degree and start a tour on adjacent vertices until you are unable to explore further. Keep deleting the edges you have traveled. Assume you end at vertex v. What we have is a trail (not path):
$$u,u_{1},u_{2}........v$$
The claim is, v has to be u.
Assume u and v are different. For every appearance of vertex v except the last one in the above list, there need to be two edges incident on v, one incoming and other outgoing. But for the last appearance, we need only one edge to enter the vertex v. Hence odd number of edges incident on v are used in the above trail. This leads to the fact there is at least one more edge on v yet to be explored. So the vertex v can not be the end of traversal. This proves u=v.
Now we have the circuit, which can be broken into disjoint edge cycles. How? For every two occurrence of a vertex y in the circuit, remove the circuit y-y from the parent circuit. Keep doing this recursively to get cycles.
Example: circuit $u,u_{1},u_{2},u_{3},u_{2},u$ be broken into $u,u_{1},u_{2},u$ and
$u_{2},u_{3},u_{2}$
Since we have deleted the edges explored, the cycles have disjoint edges. We recursively keep finding circuits in the main graph, deleting the edges until we have explored all the edges. Which completes your proof.
An easy proof- let $G_{1}$,$G_{2}$,$G_{3}$... be connected components of the graph. Since the degree of each vertex is even there are Eulerian circuit for each component. Then the trails can be broken into disjoint cycles for each component.
According to conventions that I have seen, edge and vertex connectivity for a single vertex graph is undefined.
If you try to say that the graph is vertex 1-connected or edge 1-connected, you run into an issue.
For example, if a graph is vertex 1-connected, than we can remove one vertex and disconnect the graph. However, if we remove one vertex from the graph, we get the empty graph, which is connected by convention.
Similarly, if we say that the graph is edge 1-connected, then this means that we can disconnect the graph by removing one edge. However, there are no edges to remove, so this doesn't make sense.
Best Answer
Suppose our two vertex sets have $m$ and $n$ vertices respectively; if every edge links one vertex type to another, there can be at most $mn$ edges, when every pair of different-type vertices is connected by an edge.
In this question, $m+n=6$, so $(m,n)$ can be $(1,5),(2,4),(3,3)$, up to reversal of variables. These vertex sets allow for 5, 8 and 9 edges respectively. We reject $(m,n)=(1,5)$ as we need 8 edges. The other two choices lead to the two distinct solutions to the problem (the two vertex sets are marked with white and black circles):