Prove that in a simple graph there is a path from any odd vertex to any odd vertex? 
Let $k$ be the number of vertices.
If $k = 1$, then the point is isolated and therefore has degree 0; WLOG, we can assume that no point is isolated.
With $k = 2$, there is a vertex of degree 1 to another vertex of degree 1.
With $k = 3$, suppose the graph is connected. Then 3 vertices have degree 2, and removing one edge leads to two vertices with one less degree - so the degree sequence would be [1,2,1], satisfying the hypothesis.
WLOG we can assume if $k$ is odd then the graph is not connected. Then the graph has at least one connected component. For any $k = n$ the endpoints of these connected components have degree 1.
If the graph is connected and has an even number of vertices then the vertices have degree (n-1), which is odd. QED.

Is there anything I am missing from my proof? I don't see the need to prove the case for even number of vertices with a graph that is not connected, so I didn't write it out. Also, could my proof be made more concise? Are there any errors?
 A: You seem to be proving the wrong result.  In this setting, there are two cases:


*

*All odd-degree vertices belong to the same connected component.  In this case, there is, by definition, a path from any odd-degree vertex to any other odd-degree vertex.

*Two odd degree vertices belong to disjoint components.  In this case, there are, by definition, two odd-degree vertices for which there is no path connecting them.  And this situation can occur: just take the disjoint union of any two graphs each with an odd degree vertex.
For example, there is no path connecting the two pink vertices below:

This means you're either attempting to prove something that is true by definition, or you're attempting to prove something that is false.

I suspect the correct task is to prove:

Given a connected graph with exactly two odd-degree vertices $u$ and $v$, there is an Eulerian trail from $u$ to $v$.



Since I have been asked to "Please take care in reading my proof.":

WLOG we can assume if k is odd then the graph is not connected.

Here's a connected $5$-vertex graph counterexample:



Then the graph has at least one connected component.

All graphs with at least $1$ vertex have at least one connected component.


For any k=n the endpoints of these connected components have degree 1.

Many graphs don't have "endpoints", i.e., leaf vertices (such as the one below).

If the graph is connected and has an even number of vertices then the vertices have degree (n-1), which is odd.

Here's a counterexample to this claim:

In fact, any connected subgraph of $K_n$, except $K_n$ itself, would be a counterexample.
A: There can be two cases for simple graph:
Case 1: Simple graph is not a cyclic graph so, u != v
then,
we will proof by contradiction:
Assumption To Contradict: a path starting from odd degree vertex ends at even degree
Now, take a connected component of graph, where sum of degrees = 2(#edges)
therefore,
there must be atleast one vertex of odd degree in connected component of graph.
And, Since it's a connected component there for every pair of vertices in component.
So, there is path from a vertex of odd degree to another vertex of odd degree.
Case 2: Case 1: Simple graph is a cyclic graph so, u = v 
Now if u is odd degree vertex, then v is also. Therefore there is a path for odd degree vertex to itself.
Prooved !!
A: I hope even this argument works.
Start with vertex with odd degree. It must have an edge. Traverse through it.There are two cases:
1. the other vertex have odd degree which means our problem is done.
2. the other vertex have even degree. Since it is simple graph, one edge is included with our odd degree vertex, there must be atleast one more edge to some other vertex. 
We can continue our argument recursively. As we cannot end at vertex with even degree, we must surely get one vertex of odd degree.
Please comment if there is some fault in my argument.
