Elementary proof of the fact that any orientable 3-manifold is parallelizable A parallelizable manifold $M$ is a smooth manifold such that there exist smooth vector fields $V_1,...,V_n$ where $n$ is the dimension of $M$, such that at any point $p\in M$, the tangent vectors $V_1(p),...,V_n(p)$ provide a basis for the tangent space at $p$. Equivalently, a manifold is parallelizable if its tangent bundle is trivial.
There is a theorem that states that any compact orientable 3-manifold is parallelizable, and 
there is a proof of this result which uses $spin^c$ structures and the Steifel-Whitney class.
I am wondering whether there exists a more elementary, perhaps more straightforward proof. Otherwise, I would be grateful for some intuition on why this is true.
Also, it the theorem still true without the compactness assumption? If so, is there a relatively simple proof in that case?

Here is (in my estimate) the approach outlined in Milnor-Stasheff. Let $M$ be a closed, orientable 3-fold, $w_i$ denote the $i$th Whitney class, and $o_i$ the $i$th obstruction class as it is defined in Steenrod's Topology of Fibre Bundles . As $M$ is orientable, $w_1=0$ and a straightforward calculation with Wu's formula gives $w_2$ and $w_3=0$ as well. 
From a remark in Ch.12 of Milnor-Stasheff, we have $o_2$ vanishes. So there exists a 2-frame over the 2-skeleton of $M$ (thinking of $M$ as a 3-dim CW-complex). As $\pi_2(V_2(\Bbb R^3))$ is trivial by some basic homotopy theory, the 2-frame extends over $M$ (one just takes a null-homotopy of the map on the boundary to be the map on the 3-cells). 
Now we have a trivial 2-plane bundle inside of $TM$. By endowing $TM$ with a metric we can find a complement for said 2-plane bundle, which is orientable hence trivial. So $TM$ is the Whitney sum of trivial bundles hence trivial.
