 Solution

No. Distinct singular values do not force uniqueness in the spectral norm. For an explicit example takeThe singular values of $A$ are $2>1>0$. Every $B_t$ has matrix rank one, and its error in the spectral norm isThus there are infinitely many best rank-one approximations even with a strict gap between the singular values.

The same mechanism works for every $1\le r<n$ under the question's positive singular values. Starting with $A_r$, defineWhen the singular values are distinct, ${\sigma }_1+t\ge {\sigma }_1-{\sigma }_{r+1}>0$, so every $B_t$ has matrix rank $r$, whileThe spectral norm measures only the largest remaining singular value, leaving room to change retained components without increasing that maximum. This is the nonuniqueness of spectral-norm low-rank approximation. The exceptional case $r=0$ is unique for the elementary reason given above.