Let $H=L^2(X)$ for some compact space $X$. We denote by $U(H)$ the space of unitary operators on $H$ with the topology induced by the operator norm.
Given a continuous map $t\mapsto S_t$ from $[0,1]$ to $U(H)$. Is there a continuous (or at least Borel) maps $t\mapsto \lambda_t$ and $t\mapsto f_t$ the first with values in $S^1$ and the second with values in $H\backslash\{0\}$ such that $S_t f_t =\lambda_t f_t$?
A unitary operator need not have any eigenvectors at all. For instance, let $X=S^1$ with its usual measure, and let $S\in U(H)$ be defined by $S(f)(z)=zf(z)$ (so, $S$ is the multiplication operator for the inclusion function $S^1\to \mathbb{C}$). Then if $f\in H$ is such that $S(f)=\lambda f$ for some scalar $f$, this means $zf(z)=\lambda f(z)$ almost everywhere. But this implies $f(z)=0$ almost everywhere (since if $f(z)\neq 0$ and $zf(z)=\lambda f(z)$ we can only have $z=\lambda$), so $f=0$ in $H$.
So, if $S_t$ is this $S$ for all $t$, then your desired $f_t$ and $\lambda_t$ do not exist.