Problem 25 (c) Kosniowski. A subset homeomorphic to the open disc is not a open neighbourhood of any of its points. 25.1 (c) Prove that if a subset $W$ of $\mathbb{R}^3$ is homeomorphic to the open disc $\mathring{D}^2$ then $W$ is not an open neighbourhood in $\mathbb{R}^3$ of any of its points .
Hint: If $W$ is an open neighbourhood in $\mathbb{R}^3$ of $w\in W$ then there is a subset $U_1\subseteq W$ with $w\in U_1$ and $U_1\cong \mathring{D}^3$ y definition.
Hi. I cannot understand this problem. It corresponds to the section on fundamental groups.
The idea I have is to calculate the fundamental group of $W$ which is isomorphic to the fundamental group of the open disk $\mathring{D}^2$ and also calculate the fundamental group of $U_1$ which is isomorphic to the fundamental group of $\mathring{D}^3$ and then obtain a contradiction but I don't know if it is a good way.
Assuming the hint. What would be the idea of ​​contradiction? (I ask to try to prove it)
Actualization 1. Assuming the hint. I have the following idea:
exists $\phi:U_1\to\mathring{D}^3$  homeomorphism
exists $\psi:W\to\mathring{D}^2$ homeomorphism then $\left.\psi\right|_{U_1}:U_1\to \psi(U_1)$ is a homeomorphism.
Now, $\left.\psi\right|_{U_1}\circ\phi^{-1}:\mathring{D}^3\to \psi(U_1)\subset \mathring{D}^2$ is a homeomorphism and I don't know how to conclude anymore
 A: Building on @Daniel Apsley's interesting comment, we may consider the following.
Suppose $\phi:W\to B^2$ is the homeomorphism, then $\phi|_{U_1-\{w\}}$ is a homeomoprhism, too. By the assumption (the hint), $\phi(U_1)\approx B^3$, so $\pi_1(\phi(U_1)-\phi(\{w\}))\cong\pi_1(B^3-\{0\})\cong\{1\}\implies$ simply connected.
On the other hand, we have $\phi(U_1)-\phi(\{w\})\subset B^2\subset\Bbb{R}^2$. This space cannot be simply connected (if you've seen a proof, then this is immediate)
If $\phi(U_1)-\phi(\{w\})$ were simply connected, then every map $f:S^1\to \phi(U_1)-\phi(\{w\})$ can be extended to $\tilde{f}:D^2\to\phi(U_1)-\phi(\{w\})$, so it suffices to find one counterexample to this statement. Actually, there is a simple and natural one, just have a try before revealing the spoiler. ;)

 Consider the map \begin{align}f:S^1\to\phi(U_1)-\phi(\{w\}),\\ \begin{pmatrix}\cos(\theta)\\\sin(\theta)\end{pmatrix}\mapsto \phi(w)+\begin{pmatrix}\epsilon\cos(\theta)\\\epsilon\sin(\theta)\end{pmatrix}\end{align} where $\epsilon$ is carefully chosen so that the radius fits in the target space. Then, one may notice that there exists another map \begin{align}f^{-1}:y\mapsto \dfrac{y-\phi(w)}{||y-\phi(w)||}\end{align} that acts as an inverse since $(f^{-1}\circ f)(x)=x$ by direct computation. Then, the continuous map $f^{-1}\circ \tilde{f}$ induces $f^{-1}_*\circ\tilde{f}_*$, which $\twoheadrightarrow$. But this is impossible because $f^{-1}_*\circ\tilde{f}_*:\pi_1(D^2)\to\pi_1(S^1)\not\cong\{1\}$.

