Why does the subgraph remain connected at each stage of Fleury's algorithm? On pages 42-43 in [1], it says:

We conclude our introduction to Eulerian graphs with an algorithm for constructing an Eulerian trail in a give Eulerian graph. The method is know as Fleury's algorithm.
THEOREM 2.12 Let $G$ be an Eulerian graph. Then the following construction is always possible, and produces an Eulerian trail of $G$.
Start at any vertex $u$ and traverse the edges in an arbitrary manner, subject only to the following rules:
(i) erase the edges as they are traversed, and if any isolated vertices result, erase them too;
(ii) at each stage, use a bridge only if there is no alternative.
Proof. We show first that the construction can be carried out at each stage.
Suppose that we have just reached a vertex $v$, erasing the edges as we go. If $v \neq u$, then the subgraph $H$ that remains is connected and has only two vertices of odd degrees, $u$ and $v$. ...

Why is $H$ connected? I can't prove it. Does anyone has any idea? Thanks in advance.
[1] Robin J. Wilson, Introduction to Graph Theory, 5th ed., Prentice Hall, 2012.

A "bridge" is an edge which when removed disconnects the graph. You never cross one until forced. Until then, $H$ remains connected simply by this choice. 
So consider when you are forced to cross the bridge. Why are you required to cross the bridge? Because there are no longer any other edges connecting to this vertex. What happens when you cross the bridge? The graph becomes disconnected. It now has two pieces. But one piece is that lone vertex you just left, which is then removed since it is now isolated. Now only the other piece is left. Therefore the remaining graph is still connected.
