In the positive orthant you may use the following argument.
We have $\sin \angle(u,v)=\frac{\|u\times v\|}{\|u\|\cdot \|v\|}$, where $\times$ denotes the vector product. Note that the coordinates of $u\times v$ are exactly $\pm \|u_{\Omega}\times v_{\Omega}\|$. Combined with Pythagoras and obvious inequalities like $\|u_{\Omega}\|\leqslant \|u\|$ we get $\sin^2 \angle(u,v)\leqslant \sum_{\Omega} \sin^2 \angle(u_{\Omega},v_{\Omega})$. Assume that however $\sum \angle(u_{\Omega},v_{\Omega})<\angle (u,v)$. Note that whenever $0\leqslant x,y, x+y\leqslant \pi/2$ we have $\sin^2(x+y)=(\sin x\cos y+\cos x\sin y)^2= \sin^2 x+\sin^2 y+2\sin x\sin y \cos(x+y)\geqslant \sin^2 x+\sin^2 y.$
Applying this twice we get $\sin^2 \sum \angle(u_{\Omega},v_{\Omega})\geqslant \sum \sin^2 \angle(u_{\Omega},v_{\Omega})\geqslant \sin^2 \angle(u,v)$, a contradiction.
ED by Ju: The result can be generalized to $\mathbb R^n_{++}$ using the Lagrange's identity, as discussed in the comments below.
