Argument of Feynman for equivalence of dot product definitions

This does seem to be a gap in the argument. Maybe Feynman filled in the gap elsewhere, or maybe it's a true gap. He's a physicist so he's not aiming for full mathematical rigor, he just wants great insight.

Here's how I would fill in the gap. Suppose that $\beta = (u_1, u_2, u_3)$ is an orthonormal basis for $\mathbb R^3$. Let $Q = \begin{bmatrix} u_1 & u_2 & u_3 \end{bmatrix}$ (so the $i$th column of $Q$ is the column vector $u_i$). The change of basis matrix from the standard basis to $\beta$ is $Q^{-1} = Q^T$.

Now suppose that $x$ and $y$ are vectors in $\mathbb R^3$. Notice that \begin{align} (Q^T x) \cdot (Q^T y ) &= (Q^T x)^T Q^T y \\ &= x^T Q Q^T y \\ &= x^T y \\ &= x \cdot y. \end{align} This shows that changing basis from the standard basis to the basis $\beta$ does not change the dot product.