A ball in a particular open set. Let $U\subseteq\mathbb{R}^n$ be an open set. Let $\varepsilon >0$ and define the following open set $$U_\varepsilon=\left\{x\in U\;|\; \text{dist}(x,\partial U)>\varepsilon \right\}.$$
Now, let $x$ be a point of $U_{\varepsilon}$ and consider the closed ball $\overline{B}(x,\varepsilon)$.

Question 1. Is it true or false that $$\overline{B}(x,\varepsilon)\subseteq U_\varepsilon$$


Question 2.
Is it true or false that $$\overline{B}(x,\varepsilon)\subseteq U$$

Thanks in advance!
 A: $\overline B(x,{\varepsilon})$ might or might not be a subset of $U_{\varepsilon}.$
Example 1. Let $n=1$ and $U=(0,6)$ and let $\varepsilon=2$ and let $x=3.$ Then $\partial U=\{0,6\}$ and $\overline B(x,{\varepsilon})=[1,5]$ but $U_{\varepsilon}=(2,4).$
Example 2. Let $n=1$ and $U=(0,6)$ and let $\varepsilon=1$ and let $x=3.$ Then $\partial U=\{0,6\}$ and $\overline B(x,{\varepsilon})=[2,4] \subset (1,5)=U_{\varepsilon}.$
A: First, note that by the definition of an open set, $\partial U$ is a set of points that are not in , but that can be approached arbitrarily closely by points in . In other words, for any point $y\in\partial U$, there exists a sequence of points $\{x_n\} \subset U$ such that the distance between $$ and $_$ goes to zero as  goes to infinity.
Let's consider a point $x \in U_\epsilon$. By definition, this means that the distance between $x$ and $\partial U$ is greater than $\epsilon$. In other words, for any point $y \in \partial U$, the distance between $$ and $$ is greater than $\epsilon$.
Consider the closed ball $(,)$. This is the set of all points that are a distance of at most  from . Since the distance between  and any point $y \in \partial U$ is greater than $\epsilon$, it follows that $(,)$ is a subset of $U$ (since no points in $(,)$ can be in $\partial U$).
What about $U_\epsilon$?
