Proposition: Let $D:=\{(x,y)|x^2+y^2\leq 1\}$ and $S^1$ is boundary of $D$. Then, for all differentiable function $f:D\to S^1$, there exist $p\in S^1$ s.t., $d_p(f|_{S^1})=0$
$f|_{S^1}(S^1)$ is compact, so I want to prove above proposition by similar way in case of $f$ is $\mathbb{R}$-valued. But I don't know how to prove it.
This follows immediately from the case of $\mathbb{R}$-valued functions, since $D$ is simply connected so $f$ lifts to the universal cover. That is, there is a differentiable map $g:D\to \mathbb{R}$ such that $f=p\circ g$ where $p:\mathbb{R}\to S^1$ is the universal covering map $p(t)=(\cos t,\sin t)$.