I'm reading the book Graphs and Their Uses which contains the following theorem and proof:
THEOREM 2.3. A connected graph with 2k odd vertices contains a family of k distinct trails which, together, traverse all edges of the graph exactly once.
PROOF. Let the odd vertices in the graph be denoted by in some order.
$a_1,a_2,\dots,a_k$ and $b_1,b_2,\dots,b_k$
When we add the $k$ edges $a_lb_l, a_2b_2 ,\dots, a_kb_k$ to the graph, all vertices become even and there is an Eulerian trail T. When these edges are dropped out again, T falls into k separate trails covering the edges in the original graph.
However this doesn't seem to make sense since in the graph whose vertices have degrees 3, 1, 1, 1 there is no way to add 2 edges in such a way that the degree of all odd vertices becomes even.
What am I missing here?