= Solution
{c}
\b[No. Distinct <singular values> do not force uniqueness in the <spectral norm>.] For an explicit example take
$$
A=\begin{pmatrix}2&0\\0&1\end{pmatrix},\qquad
B_t=\begin{pmatrix}2+t&0\\0&0\end{pmatrix},\qquad -1\leq t\leq1.
$$
The <singular values> of $A$ are $2>1>0$. Every $B_t$ has <matrix rank> one, and its error in the <spectral norm> is
$$
\boxed{\|A-B_t\|_2=\max\{|t|,1\}=1=\sigma_2(A).}
$$
Thus there are infinitely many best rank-one approximations even with a strict gap between the <singular values>.
The same mechanism works for every $1\leq r<n$ under the question's positive <singular values>. Starting with $A_r$, define
$$
B_t=A_r+t u_1v_1^T,\qquad |t|\leq\sigma_{r+1}.
$$
When the <singular values> are distinct, $\sigma_1+t\geq\sigma_1-\sigma_{r+1}>0$, so every $B_t$ has <matrix rank> $r$, while
$$
\|A-B_t\|_2=\max\{|t|,\sigma_{r+1}\}=\sigma_{r+1}.
$$
The <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.
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