Unique nearest point in epsilon neighborhood of compact real manifold?

Question

I have to proof the following assertion:

Let $X$ be a compact submanifold of $\mathbb{R}^n$ and $\mathcal{U}^\varepsilon=\{p\in\mathbb{R}^n\;:\; |p-q|<\varepsilon \text{ for some }q\in X\}$. Then there is $\varepsilon>0$ such that every point $p\in\mathcal{U}^\varepsilon$ has a unique nearest point $\pi(x)\in X$.

But is this true at all?

Consider $f(x)=x\cdot\sin(1/x)$ for $x\ne0$ and $f(0)=0$. Since $f$ is continuous, $X=\{(x,f(x)\;|\;x\in[0,1]\}$ is a compact submanifold of $\mathbb{R}^2$. Let $x_1 > x_2 > \ldots$ be all roots of $f$ in $(0,1]$. For every point $\big(\frac{x_{i+1}+x_i}{2},0\big)\in\mathbb{R}^2$ both $(x_i,0)$ and $(x_{i+1},0)$ are nearest points in $X$. And since $x_{i+1}-x_i$ tends to zero, we have such points in every $\mathcal{U}^\varepsilon$.

What is wrong with my argument?

Thanks in advance!

Answer 1

Your homework problem is about smooth submanifolds.