Why sufficient
Why “if”: First, observe the following:
Lemma 1. Assume every vertex in a multigraph has even degree. Start
at an arbitrary non-isolated vertex v0, choose an arbitrary edge {v0,v1},
then choose an arbitrary edge {v1,v2}, then an arbitrary edge {v2,v3},
etc.; every time choose an edge that has not been used yet. Then sooner
or later you will arrive back to v0 and thus get a simple circuit.
Indeed, in the above procedure, once you entered a vertex x, there will
always be another unused edge to exit x because x has an even degree
and only an even number of the edges incident with it had been used
The only edge from which you may not be able to exit after entering
it is v0 (because an odd number of edges incident with v0 have been
used as you didn’t enter it at the beginning) , but if you
have reached v0, then you have already constructed a circuit.