Elementary proof of the fact that any orientable 3-manifold is parallelizable

I don't know of a totally elementary proof of this result, but here is some context for it. More generally, an $n$-manifold $M$ (without boundary) has a classifying map $f : M \to BO(n)$ for its tangent bundle. Knowing when $M$ is parallelizable is equivalent to knowing when $f$ is null-homotopic. There is a general machine for doing this involving lifting $f$ higher and higher through the stages of the Whitehead tower of $BO(n)$, and it tells us that the complete set of obstructions to solving this problem are a set of cohomology classes in $H^k(M, \pi_k(BO(n)), k \le n$, each of which is well-defined provided that the previous one vanishes, such that $f$ is null-homotopic iff all of the classes vanish.

The construction of the first such class goes like this. Assume for simplicity that $M$ is connected. The first question is whether $f$ lifts to the universal cover of $BO(n)$, which is true iff $f$ induces the zero map on $\pi_1$ by standard covering space theory. Now, $\pi_1(BO(n)) \cong \mathbb{Z}_2$, and the induced map on $\pi_1$ gives a homomorphism $\pi_1(M) \to \mathbb{Z}_2$ which corresponds precisely to the first Stiefel-Whitney class $w_1$. This class vanishes iff $f$ lifts to the universal cover of $BO(n)$, which is $BSO(n)$, iff $M$ is orientable.

Now we want to try lifting $f$ to the $2$-connected cover of $BSO(n)$; this is analogous to the universal cover but involves killing $\pi_2(BSO(n)) \cong \mathbb{Z}_2$ (for $n \ge 3$) instead of $\pi_1$. Whether this is possible is controlled by the map

$$BSO(n) \to B^2 \mathbb{Z}_2$$

inducing an isomorphism on $\pi_2$. This is equivalently a universal characteristic class in $H^2(BSO(n), \mathbb{Z}_2)$ which turns out to be precisely the second Stiefel-Whitney class $w_2$. This class vanishes iff $f$ lifts to the $2$-connected cover of $BSO(n)$, which is $BSpin(n)$, iff $M$ has a spin structure.

The first surprise in this story is that (when $n \ge 3$) $BSpin(n)$ also turns out to be the $3$-connected cover; in other words, $\pi_3(BSpin(n)) = 0$, so the next step of this story involves $\pi_4$ and can be ignored for $3$-manifolds. If $M$ is a $3$-manifold, not necessarily compact, admitting both an orientation and a spin structure, then the classifying map of its tangent bundle lifts to a map $M \to BSpin(3)$, but since the latter is $3$-connected any such map is nullhomotopic. So:

A $3$-manifold, not necessarily closed, is parallelizable iff the first two Stiefel-Whitney classes $w_1, w_2$ vanish iff it admits an orientation and a spin structure.

To give an indication of the generality of this machinery, for $4$-manifolds the next step involves computing $\pi_4(BSpin(4)) \cong \mathbb{Z}^2$ and then lifting to the $4$-connected cover of $BSpin(4)$, which is called $BString(4)$. This says that the next obstruction to a $4$-manifold with an orientation and a spin structure being parallelizable is a pair of cohomology classes in $H^4(M, \mathbb{Z})$ which I believe turn out to be the Euler class $e$ and the first fractional Pontryagin class $\frac{p_1}{2}$ (a certain characteristic class of spin manifolds that when doubled gives the Pontryagin class $p_1$) respectively. Since $H^4(M, \mathbb{Z})$ is always torsion-free for a $4$-manifold, the conclusion is that

A $4$-manifold, not necessarily closed, is parallelizable iff the characteristic classes $w_1, w_2, e, p_1$ all vanish iff it admits an orientation, a spin structure, and a string structure.

And for $5$-manifolds and higher we really need to talk about $\frac{p_1}{2}$ and not just $p_1$.

The second surprise in this story is that for closed $3$-manifolds the condition that $w_2$ vanishes is redundant: it is implied by orientability by a standard computation with Wu classes. In other words, closed orientable $3$-manifolds automatically admit spin structures. I don't have a good intuitive explanation of this; it comes from a relationship between the Stiefel-Whitney classes, Steenrod operations, and Poincaré duality that I don't understand very well.

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.

Cf. "On some theorems of Geroch and Stiefel" by Phillip E. Parker, Journal of Mathematical Physics 25, 597 (1984); https://doi.org/10.1063/1.526209 Where "Modern proofs of two theorems of Geroch (spinor spacetimes and globally hyperbolic spacetimes are parallelizable) and a theorem of Stiefel (orientable 3-manifolds are parallelizable) are given, using the computationally efficient obstruction theory of algebraic topology. These techniques also easily show that, in fact, Geroch's second theorem is a corollary of Stiefel's theorem."