Eulerian walk proof: If a connected graph has exactly two nodes with odd degree, then it has an Eulerian walk? Prove that: If a connected graph has exactly two nodes with odd degree, then it has an Eulerian walk. Every Eulerian walk must start at one of these and end at the other one. How shall I prove this?

Let $(u, v)$ be the only two vertices of odd degree in your graph $G$. Consider $G'$ where you add an edge between $v, n$ to $G$. Then every vertex in the graph has even degree and is connected. Therefore you have a Eulerian circuit. Remove the edge and you get a Eulerian walk. Note that $G + e$ is not simple, as in, there may be more than one edge between two vertices. This is fine because Eulerian circuits still hold when we consider graphs that are not simple. 
