Why, the Valence Bundle is non-trivial and the complete Bloch Bundle isn't?

That the valence bundle is generically non-trivial requires no proof. Just think about any non-trivial example of a Chern insulator.

Possibly the easiest way to see that the Bloch bundle is trivial (without introducing a lot of notation) is to use the fact that (at least for bundles over the 2-torus) if the Chern number of the projection defining the bundle is zero then the bundle is trivial. But the formula defining the Chern number of a projection, for example, formula (3) in https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.51.51 $$ mathrm{Chern}(P) = frac{1}{2pi i}int_{mathbb{T}^2} operatorname{tr}(P(k)[(partial_{k_1}P)(k),(partial_{k_2}P)(k)])mathrm{d}k$$

If you take the Bloch bundle, that is, all the space, then the corresponding projection is simply the identity matrix $I$ for each value of $k$: $P(k)=I$. This means that its Chern number is manifestly zero, and so (over $mathbb{T}^2$) it is the trivial bundle.

The Bloch bundle is always defined to be a trivial bundle. For example, if you consider a four-band model of a 2D system, the Bloch bundle is $mathbb{T}^2timesmathbb{C}^4$. Every Bloch is described by a point $vec{k}$ in the Brillouin zone ($mathbb{T}^2$) and a length-four unit vector $vec{u}$ ($mathbb{C}^4$). We identify the points $(vec{k},vec u)sim (vec{k}+frac{2pi}{a_{x(y)}}hat{x}(hat{y}),vec u)$. This implies we are indeed a product bundle. The only time you can have nontrivial topology is if you consider a subbundle of the Bloch bundle, such as the valence (or conduction) bundle. Subbundles of trivial bundles can be trivial or twisted, depending on what subbundle you choose!

This is roughly analogous to the simpler case of manifolds. You can embed every manifold in the trivial manifold $mathbb{R}^N$, just as we are embedding the valence bundle in the trivial bundle. These manifolds can either be trivial themselves, or have non-trivial structure.