The theorem goes as follows:
Let $Y \subseteq \mathbb{R}^n$ smooth, compact manifold. Then for a small enough $ \epsilon>0 $, each $w \in Y^\epsilon := \{ w \in \mathbb{R}^n \big \vert \exists y \in Y: |y-v|< \epsilon\} $ has a unique closest point in Y, denoted $\pi(v)$. Moreover $\pi : Y^\epsilon \rightarrow Y$ is a submersion with $\pi_{|Y}=id$. Parts of the proof are left as an exercise, including the statement that $\pi(w)$ is the unique closest point
At this point I have shown that $h: N(Y) \rightarrow \mathbb{R}^n, \; h(y,v)=y+v$, where $N(Y)$ is the normal bundle, $y\in Y$ and $v\in N_yY$, maps a neighborhood of $Y \times 0 \subseteq N(Y)$ onto $Y^\epsilon\subseteq \mathbb{R}^n$, which is a neighborhood of Y. Guillemin defines $\pi =\sigma \circ h^{-1} $, where $\sigma:N(Y) \rightarrow Y$ is the natural projection.
I can show that if $\pi(w)$ is indeed a closest point, then for some $\epsilon > 0$ small enough it is unique, but what I haven't been able to show is that $|| \pi(w) -w|| $ really is minimal.
I know that a closest point must exist by compactness of $Y$, let's call it $y_{min}$, and I think I have shown that a necessary condition for $y \in Y$ to minimize $||y-w||$ is $(y-w) \perp T_yY $, which would fit my intuition. From the definition one can see that $\pi(w)-w \perp T_{\pi(w)}Y$. But how can I see that $h^{-1}(w)= (y_{min}, w-y_{min})$, such that $\pi(w)=\sigma((y_{min}, w-y_{min}))=y_{min}$ ?
I have tried so far to assume that $h^{-1}(w)= (\tilde{y}, w-\tilde{y})$ for some $y_{min}\neq \tilde{y} \in Y$ and either show that $\tilde{y}$ is another closest point, since I show uniqueness only later, or assumed that $\tilde{y}$ is not a closest point and find a contradiction to $h^{-1}$ being a diffeomorphism on $Y^\epsilon$, but neither has been successful so far. Any help would be appreciated.