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 $v$. To show that the construction can be carried out, we must show that the removal of the next edge does not disconnect $H$ $-$ or, equivalently, that $v$ is incident with at most one bridge. But if this is not the case, then there exists a bridge $vw$ such that the component $K$ of $H-vw$ containing $w$ does not contain $u$ (see the figure below). Since the vertex $w$ has odd degree in $K$, ...
Please look at the last sentence of the proof. Why does the vertex $w$ have odd degree in $K$? Thanks in advance.
[1] Robin J. Wilson, Introduction to Graph Theory, 5th ed., Prentice Hall, 2012.
Currently the vertex $w$ has even degree in $H$. This is because it had even degree in $G$, and if we have visited it $k$ times so far, we have used up $2k$ of its edges (we arrived at $w$ $k$ times, and left it $k$ times), so an even number remain. Removing the edge $vw$ leaves it with odd degree, and all the edges from $w$ apart from $vw$ lie in $K$, so it has odd degree in $K$.