2-connected graph has a strongly connected orientation 
Prove that any $2$-connected graph has a strongly connected orientation. 

I have the definition of strongly connected for directed graph. A directed graph is strongly connected if there is a path from $a$ to $b$ and from $b$ to $a$ whenever $a$ and $b$ are vertices in the graph. And we also know that a graph G is 2-connected if and only if there exists, for any two vertices of G, a cycle in G containing these two vertices.
And since a cycle exists between any two vertices, does this satisifies the condition of a strongly connected orientation? As a cycle contains any x and y in the graph implies that there is a path from x to y, and a path from y to x. 
I have not formally written my proof yet, it is just a draft and brainstorming at the moment. would you please point out if I am on the right track? 
 A: You can apply Robbins's theorem stated below to prove it.

Bobbins's Theorem: An undirected graph $G$ can be oriented to obtain a strongly-connected directed graph if and only if
  G is $2$-edge-connected.

You can find a proof for it in this lecture note. Here, "$2$-edge-connected" means removing any one edge from $G$ does not diconnect $G$ and it is not same as "$2$-connected" since "$2$-connected" usually denotes "$2$-vertex-connected".  However, it can be easily proved that

Lemma: If an undirected graph $G$ is $2$-connected, it is also $2$-edge-connected.

A: Alternate approach.  Prove the result for multigraphs where 2 nodes can have multiple edges.  See https://en.wikipedia.org/wiki/Multigraph.  (The type of multigraph that I am thinking of has multiple labeled edges, loops allowed.)
The trick is that you find a cycle, label all of its edges with an orientation, drop them and then do https://en.wikipedia.org/wiki/Edge_contraction#Vertex_identification of the whole cycle to get a multigraph with fewer vertices.  Then you have to prove that a strong orientation on that contracted multigraph can be lifted back to a strong orientation on the original multigraph.
With this step, you can prove the result by strong induction, with a base case of a single vertex with an arbitrary number of connected loops.  (Any orientation in this case is a strong orientation.)
This will prove it for multigraphs.  Simple graphs are, of course, a special case of multigraphs.
