I don’t think this is true. For example $H$ could be just the singleton space $\{0\}$, which is not homeomorphic to the unit ball…

In the way it is written it $\epsilon$ seems to be a chosen constant. So choosing $\epsilon = 1$ and $x=0$ we obtain $H$ as a special instance of $H_\epsilon(x)$. And the assumptions on $H$ do not suffice to make the desired statement true.

@XanderHenderson Assuming we are working with a star-convex : How would one show that open and star convex is homeomorphic to the unit ball ? I understand that this works for open and convex, but star-conex doesn't imply being convex. mathworld.wolfram.com/StarConvex.html

I think you are missing the assumption that $\cal H$ is an open set.

This would be true if we assumed $\mathcal{H}$ is open though? @Surb

@Seabourn I misread your original statement---I had assumed that you were describing an $\epsilon$-neighborhood of $\mathcal{H}$, which is a set $$ \mathcal{H}_{\varepsilon} = \bigcup_{h\in\mathcal{H}} B(h,\varepsilon). $$ Because $\mathcal{H}$ is star-convex, so too is $\mathcal{H}_{\varepsilon}$. Then the argument here applies. But that is not the set you are working with so, my previous comments were not relevant.

@Surb That doesn't seem to meet the definition of $\mathcal{H}$. My understanding of the definition is that if $h\in\mathcal{H}$, then the segment joining $h$ to the origin is also in $\mathcal{H}$. Your set does not meet that definition (e.g. the point $(\varphi, 0)$ is on the segment joining $0$ to $(2,0)$, but is not in $\mathcal{H}$, where $\varphi$ is the golden ratio).

In any event, as stated, the result is not true. Can you please check the statement of the claim, and make sure that you have copied it down correctly? Can you please also add a little context: what are you studying, and at what level? My guess is that invoking results about star-shaped domains is a little off-topic, but I don't know what tools you are expected to use. Your question needs some clarity...

@XanderHenderson Thanks for your help. The only thing I was missing is the assumption that $\cal{H}$ is open. The rest is all likewise the exercise sheet I have.

@XanderHenderson The set in my now deleted comment was containing the points you mentioned, for recall, I suggested to consider $H = \{t(x,y)\mid t\in[0,1], x,y\in \Bbb Q, \|(x,y)\|=2\} \cup \{(x,y)\mid x,y\in \Bbb R, \|(x,y)\|\leq 1\}$. I deleted the comment because this set is not open.

It is true, but not easy. See math.stackexchange.com/q/184453.

Thanks for your help @PaulFrost ! Just one further question: the fact that a star-shaped set is diffeomorphic to $\Bbb R^n$ implies it is homeomorphic to the unit ball $K$ ?

I understand that part, I meant more the partof being homeomorphic to $\Bbb R^n$ implies being homeomorphic with the unit ball $K$. Thanks for your help @XanderHenderson

There are a lot of ways to show that $\mathbb{R}^n$ is homeomorphic to the unit ball. The usual approach is to use the arctangent function.

Well, I agree with that. But looking at this (proofwiki.org/wiki/Composite_of_Homeomorphisms_is_Homeomorphism) and looking at the proofs abot the two topics in discussion, we define a function from a starshaped set $\omega$ to $\Bbb R^n$ (let $f: \omega \rightarrow \Bbb R^n$) and to prove the second topic, we define $g : K \rightarrow \Bbb R^n$. This doesn't satisfy the link I just posted, about the preservation of homeomorphisms through composition of functions. How would one proceed? @XanderHenderson

@Seabourn Take $g : \mathbb R^n \to K, g(x) = \frac{x}{1 + \lVert x \rVert}$.