It says for each $t\in S^1$ there is a $v_t$ such that $\text{dist}(0,\text{sp}(c(s))\backslash \{0\})\geq d_t>0$, which is not true. A counter example is $c(e^{it})=\text{diag}(1/2,|t|)$. When $e^{is}$ is near $1$, $\text{sp}(c(e^{is}))=\left\{\begin{array}{}\{1/2\}&,s=0\\ \{1/2,|s|\}&,s\neq 0\end{array}\right.$, therefore $\inf_{e^{is}\in v_1}\text{dist}(0,\text{sp}(c(e^{is}))\backslash \{0\})=0 $.
However there is a $f_t$ such that $f_1(c)(x)=q_1(x)$ is a projection of rank at least $1$ in $v_1$, if $v_1$ is sufficiently small. But then $\|q_i(s)-q_{i+1}(s)\|<1/8$ fails, because they may have different ranks.