Let $E\subset\mathbb{R^{n}}$ and $\hat{p}\in\mathbb{R^{n}}$, then all interior points of $E$ are limit points of $E$ as well. Let $E\subset\mathbb{R^{n}}$ and $\hat{p}\in\mathbb{R^{n}}$, then all interior points of $E$ are limit points of $E$ as well.
If $\hat{p}\in int(E)$ where $int(E)$ is the set containing all the interior points of $E$, then there is an $r'>0$ such that an open ball centered at $\hat{p}$ with an $r'$ radius and completely contained in $E$ can be formed.
My intention is to prove that $\forall r>0$, $(Ḅ̣_{r}(\hat{p})\setminus \{\hat{p}\})\cap E \neq \emptyset$. Where $Ḅ̣_{r}(\hat{p})$ is the set including all points contained in the open ball of radius $r$ (in other words: $Ḅ̣_{r}(\hat{p})=\{\hat{s}\in \mathbb{R^{n}}| \thinspace d(\hat{p},\hat{s})<r\}$).
I was thinking that perhaps the way to go is to take an arbitrary $\hat{p}\in int(E)$ and an $R$ such that $R=min\{r,r'\}$ and then try and get to the definition of a limit point from there; however I seem to be stuck and I'm starting to wonder if this is the way to go at all.

Hint: If $\hat{p}\in{\rm int}E$ then there exists $Ḅ̣_{r}(\hat{p})$ with
$Ḅ̣_{r}(\hat{p})\subset E$.
What do you add for the punctured ball
$Ḅ̣_{r}(\hat{p})\smallsetminus\{\hat{p}\}$?
What happens if
$(Ḅ̣_{r}(\hat{p})\smallsetminus\{\hat{p}\})\cap E=\varnothing?$
