Characteristics of Eulerian Graph/Sufficient Condition/Proof 2

Proof
Suppose that an (undirected) graph $G$ is connected and its vertices all have even degree.

From Graph with Even Vertices Partitions into Cycles, we can split $G$ into a number of cycles $\mathbb S = C_1, C_2, \ldots, C_k$.

Start at any vertex $v$ on cycle $C_1$ and traverse its edges until we encounter a vertex of another cycle of $\mathbb S$, $C_2$ say.

The edges of $C_2$ are then traversed, and then the traversal of $C_1$ is resumed when it is returned to.

As $C_1$ is traversed, the journey is interrupted so as to traverse any other cycles of $\mathbb S$ in the same way that $C_2$ is traversed.

Eventually the beginning of $C_1$ is reached, which is vertex $v$.

Thus there exists a circuit which includes $C_1$ and at least one other cycle (unless $C_1$ is the only cycle), as $G$ is connected.

If this circuit contains all the cycles $C_1, C_2, \ldots, C_k$, the required Eulerian circuit has been found.

If not, then the circuit just generated is traversed.

Because $G$ is connected, other cycles in $\mathbb S$ will be encountered.

These, again, are traversed as they are encountered.

This process is continued till all the cycles have been included in the circuit.

At this stage, the required Eulerian circuit has been found.

Hence $G$ is Eulerian.