I'd love to see a direct proof using more elementary methods.

The proof below appeal to a famous result of Schoenberg (I've simplified the statement a bit), and basic linear algebra.

> **Schoenberg's theorem** (see e.g., [Prop. 3.2, 1]).  Let $X$ be a nonempty set and $\psi: X \times X \mapsto \mathbb{R}$ be positive definite kernel. Then, there exists an RKHS $H$ and a map $\varphi : X \to H$ such that
\begin{equation*}
  \|\varphi(x)-\varphi(y)\|_H^2 = \frac{1}{2}[\psi(x,x)+\psi(y,y)] - \psi(x,y).
\end{equation*}

We show that the function $\psi(A,B) = \det(A^TB)$ is positive definite, which as a result of Schoenberg's theorem shows that
\begin{equation*}
  1-\det(A^TB) = \|\varphi(A)-\varphi(B)\|_H^2,
\end{equation*}
from which the triangle inequality is immediate. 

To prove the positive definiteness of $\psi$, we show that it is an inner-product by invoking the [Cauchy-Binet formula][1] (using Wikipedia's notation, except that for us $A$ is $m \times n$):
\begin{equation*}
  \det(A^TB) = \sum_{S \in \binom{[m]}{n}} \det(A^T_{[n],S})\det(B_{S,[n]}) = \sum_{S \in \binom{[m]}{n}} \det(A_{S,[n]})\det(B_{S,[n]}) = \langle \phi(A), \phi(B)\rangle.
\end{equation*}

  **[1]**  C. Berg, J. P. R. Christensen, and P. Ressel. *Harmonic Analysis on Semigroups: Theory of Positive Definite and Related Functions*, Springer GTM 100, 1984.


  [1]: http://en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula