Prove that for any connected graph, there exists a closed walk that visits every edge exactly twice. For this to occur, every vertex has to have an even degree. Since this is a connected graph, we know that this must be true. 
How do you prove that the closed walk visits every edge exactly twice? Does the existences of a closed walk mean that there is a cycle in the graph? 
 A: @Daniel Schepler gives an excellent answer in the comment above. Here is a constructive proof.
If a graph is connected it contains a spanning tree. On that spanning tree, you can find a walk that visits each edge exactly twice: draw the spanning tree as a planar graph and walk around it always keeping an edge on your right.
Each tree-edge is visited exactly twice.
Now add the missing edges one by one. Each time you add an edge, use it to add a detour (back and forth) to the walk.
Each non-tree-edge is visited exactly twice.
A: Proof by induction on the size (number of edges) of the (finite) connected graph $G.$
Let $G$ be a connected graph of size $m.$ Assume that every connected graph of size less than $m$ has a closed walk that traverses each edge twice. If $m=0$ then $G$ consists of a single vertex and the trivial walk does the job. Suppose $m\gt0.$ Choose an edge $uv$ of $G.$
Case 1. $G-uv$ is connected.
By the inductive hypothesis, $G-uv$ has a closed walk $W$ that traverses each edge twice. Start at $u;$ walk on $uv$ from $u$ to $v;$ follow the closed walk $W,$ beginning and ending at $v;$ walk back to $u$ on the edge $uv.$
Case 2. $G-uv$ is disconnected.
Let $H_u$ be the component of $G-uv$ containing $u,$ and let $H_v$ be the component containing $v.$ By the inductive hypothesis, $H_u$ has a walk $W_u$ that traverses each edge twie, and $H_v$ has a walk $W_v$ that traverses each edge twice. Start at $u;$ follow the walk $W_u$ from $u$ back to $u;$ walk on $uv$ from $u$ to $v;$ follow $H_v$ from $v$ back to $v;$ return to $u$ on the edge $uv.$
