Proof that if a simple Graph contains at most two nodes with odd degree then it has a Euler walk My proof would be start as the following :
In general if there are two node at most, then one node used to start walking and the other to end.
A) If we start from odd one, this means we have two scenarios:
1) if odd =1 then we start from it and leave it   forever; this means: visiting once (Starting Point).
2)if odd>1 then we have to revisit it again, but we will leave it because # edges will enforce us to leave it at the end.
B) if then having another node with odd degree, this mean we have to stop at it. because entering node with odd edges enforces us to stay on it at the end, being no possibility to go out forever.
I think this proofs that Lemma ? Is not it ?
please see the picture bellow:
http://i.stack.imgur.com/cCnqS.jpg
 A: If there is exactly one vertex of odd degree there is no Euler walk, what you say is not possible, there must always be an end vertex. It is however true that no graph has exactly one vertex of odd degree this is becase the number of vertices of odd degree is always even. This can be proved by noticing the sum of the degree of the vertices is twice the number of edges and hence even.
The theorem you should prove is that an Eulerian circuit for a connected graph $H$ (closed walk) exists if and only if all vertices have even order. We prove this theorem by strong induction on the number of edges of $H$. When $H$ has three edges the graph is a cycle and is therefore true.
Let $H$ be a graph in which every vertex has even degree. Then $H$ contains a circuit $C$. After we remove that circuit $C$ (only the edges) we are left with a graph $H'$ with a collection of connected components (possibly  only one), each of which also has only vertices of even degree, and by the inductive hypothesis each component has its own eulerian circuit. We now build an eulerian circuit for $H$. pick a vertex $w$ on $C$, start moving along the circuit $C$ , each time we reach a vertex $v$ in $C$ check to which connected component of $H'$ the vertex $v$ belongs, if it is the first vertex we have visited of that component, then we move along the eulerian circuit of that connected component until we come back to the same vertex of $C$, If $v$ is not the first vertex we reach of that connected component then simply keep moving along $C$. Do this until we get back to the initial vertex $w$. This closed walk is the eulerian circuit we required.
This theorem directly implies if $H$ has only vertices of even degree then it has an eulerian walk (since Eulerian circuits are a special type of Eulerian walk). We can use this to prove the case in which there are exactly two vertices of odd degree in $H$. Let $u$ and $v$ be the vertices. If $uv$ is an edge in $G$ create a new graph $H'$ by removing edge $uv$. Let $u,x_1,x2\dots u$ be an eulerian circuit for $H'$, then $u,x_1,x_2\dots u,v$ is an eulerian walk for $H$.
If $uv$ is not an edge of $H$ create $H'$ by adding the edge $uv$. $H'$ has only vertices of even degree and hence it has an eulerian circuit. Let $u,x_1,x_2,\dots ,vu$ be the circuit, then $u,x_1,x_2\dots v$ is an eulerian walk for $H$.
A: Let $G$ be a connected graph such that every vertex has even degree except for at most 2 vertices.  Then, there are either 0 vertices of odd degree, in which case it contains an eulerian circuit (and so an eulerian trail), there is 1 vertex of odd degree (as we know from the handshake lemma, this is impossible), or 2 vertices of odd degree.  Thus, we may assume there are exactly 2 vertices of odd degree, say $u,v$.  
Case 1:  $uv \in E(G)$:  In this case, we remove the edge $uv$ to arrive at a graph, $G'$, in which every vertex has even degree and thus $G'$ is eulerian,  we need to add our edge $uv$ back in and find an eulerian trail.  I will leave this up to you!
Case 2:  $uv \notin E(G)$:  In this case we may add the edge $uv=$ to form a graph $G'$ in which every vertex has even degree.  Thus, $G'$ contains an eulerian circuit say $(e_1e_2e_3...e_m)$  where $e_m$ and $e_1$ both share a vertex.  Then, by definition, $uv=e_i$ for some $i$ in the eulerian circuit.  Remove this edge to arrive back at $G$, our eulerian circuit is destroyed in the process, but an eulerian path remains.  Remove $e_i$ from the eulerian circuit to arrive at an eulerian path $e_{i+1}e_{i+2}...e_me_1e_2...e_{i-1}$ which begins on one of $u,v$ and ends on the other.
A: To prove: "A graph that has at most two nodes with an odd degree has an Eulerian trail."
(A trail is a walk that visits each edge at most once. An Eulerian trail is a trail that visits every edge in the graph exactly once.)
Jorge has shown in his answer that "a graph has an Eulerian circuit if and only if all its vertices have even degrees."
The proof we need is a corollary to this theorem.
Notice that the number of vertices with an odd degree must be even in any graph. So our graph can have either 0 or 2 such vertices. If it has 0 such vertices, then there is an Eulerian circuit by the above theorem. An Eulerian circuit is also an Eulerian trail, hence the proof is complete.
Let G be our graph with vertices u and v having an odd degree each. We add an edge e that connects u and v to form the graph H. All vertices of H have an even degree. By the above theorem, it has an Eulerian circuit. This Eulerian circuit, by definition, visits all edges of H including e. If we remove the edge e, this Eulerian circuit reduces to a walk with endpoints u and v in the graph G. This is in fact the Eulerian trail that visits each edge in G exactly once.
A: Let We have a Graph G with vertices set V(G).let u and v be two vertices with odd degree.
For every times, We entering u and leaving u, it requires two degree.
Our Eulerian Trail will look like u,u1,u2,....,v2,v1,v. (Starting from u and ending at v).
Let we start from vertex u,
For traversing k times, it requires 2k degree,where k=1,2,3..
but we have odd degree of u. So deg(u)=2k+1, means we have to leave this vertex after traversing k times.
Same case of vertex V. Means after traversing v, k times, It also requires 2k+1 degree.
So, Finally we will stop at v.
Hence, It states that, A graph G has Eulerian trail, iff it has atmost two vertices of odd degree..
I hope you all get it :)..
A: First note that the graph in question has either $2$ odd-degree nodes, or zero. By the handshaking lemma, the sum of vertex degrees is twice the number of edges, so is even.
For a graph with two odd-degree vertices we take a first step of identifying a path between those two vertices. Removing those edges leaves the graph with only even-degree vertices, since the end nodes of the path ae reduced in degree by $1$ and all other nodes on the path by $2$. It is possible that the graph is now not connected, but this is not a problem.
Now we iteratively find cycles among the reducing edge count:

*

*Choosing a vertex that already been affected previously (or an arbitrary vertex if this is the first step), remove one edge. This does not further disconnect the component it is in, because that would leave one odd-degree node in each resulting component, impossible (handshaking applies to each component).

*Now find a path between the two vertices associated with the removed edge, and removed that - together with the initial edge, this makes a cycle. Again all the resulting reduced degree values are even.

Then the trail can be formed by the initial path (or cycle), with the generated cycles added at each affected vertex along the way.
By way of a very simple illustration, in the graph below, the red edges denote the initial trail between odd-degree nodes, then adding a blue cycle that empties one component, then a green cycle in the other component, then an orange/yellow cycle off the green cycle (or off the original trail) to finish. The Eulerian trail would go round the blue cycle, step once along the red, go round some of the green cycle, go round the orange cycle, finish the green cycle and finish the red trail.

