If you have unit vectors $u$ and $v$, which are orthogonal to each other, in $n$ dimensional space, is it not true to say
- they lie in a plane
- viewing the 2d projection of that plane, the axis of your current coordinate system could be redefined in such a way that vectors $u$ and $v$ become $(1,0,0...)$ and $(0,1,0...)$ respectively or some analogous form where they lie completely on one of the newly defined axis.
And so, if the above is correct, you would then be correct in saying that $||u-v|| = \sqrt{2}$, for any two orthogonal unit vectors in n dimensional space, because the simple case $||(1,-1,0...)|| = \sqrt{2}$ and the vectors can always be manipulated into a form that makes it true.
First of all, $$|| u - v || = \sqrt{<u-v,u-v>} = \sqrt{||u|| - 2u \cdot v + ||v||} = \sqrt{1 + 0 + 1} = \sqrt2.$$
Orthogonal vectors are linearly independent. Two linearly independent vectors always lie in a plane, namely the plane spanned by them. Also you can send any two linearly independent vectors to any arbitrary vectors. But in order for length to be preserved, this mapping needs to be orthogonal. So $(2,0,0)^t$ and $(0, 2, 0)^t$ are orthogonal, but their difference does not have length $\sqrt2$, because of scaling, which is not an orthogonal mapping. But given two sets of linearly independent orthogonal unit vectors, you can always find an orthogonal mapping between them. And so your reasoning works. But the more straightforward approach is what I have done above.