To prove or disprove that for each $y\in N_\epsilon(A)\setminus A$, $B(y,\epsilon)\cap [X\setminus N_\epsilon(A)]\neq \emptyset$. Let $A$ be a subset in a normed space $(X,\|\cdot\|)$, and $N_\epsilon(A)=\bigcup\limits_{a\in A}B(a,\epsilon)$, where $B(a,\epsilon)$ is open ball centered at $a$ with radius $\epsilon>0$. To prove or disprove that for each $y\in N_\epsilon(A)\setminus A$, $B(y,\epsilon)\cap [X\setminus N_\epsilon(A)]\neq \emptyset$, if $X\setminus N_\epsilon(A)\neq\emptyset$.
What I tried is:
I wanted to prove that if $B(y,\epsilon)\cap [X\setminus N_\epsilon(A)]= \emptyset$, then $y$ has to be in $A$, but could not succeed. We have, for two distinct points $u,v\in X$, there is an isometry $$I:[0,\|u-v\|]\to \{\alpha u+(1-\alpha)v:0\leq \alpha \leq 1\}$$ such that $I(0)=u, I(\|u-v\|)=v, I([0,\|u-v\|])=\{\alpha u+(1-\alpha)v:0\leq \alpha \leq 1\}$. I took a $z\in X\setminus N_\epsilon(A)$, so $[0,\|y-z\|]$ is isometric to $\{\alpha y+(1-\alpha)z:0\leq \alpha \leq 1\} $. Then, if $\epsilon\geq\|y-z\|$, then done; but if $0<\epsilon<\|y-v\|$, there is $w\in \{\alpha y+(1-\alpha)v:0\leq \alpha \leq 1\}$, such that $I(\epsilon)=w$, then how to prove $w\in X\setminus N_\epsilon(A)$.

This is not true: let $\varepsilon=1$ and in the complex plane take $A$ to be a disk minus a disk, $A=\{z\in\mathbb{C}: 0.9<|z|<3\}$. Then $0\not\in A$, but $0\in N_\varepsilon(A)$. On the other hand $N_\varepsilon(A)=\{z\in\mathbb{C}:|z|<4\} \neq\mathbb{C}$, so obviously $\mathbb{C}\setminus N_\varepsilon(A)\neq\emptyset$, and $D(0,\varepsilon)\cap (\mathbb{C}\setminus N_\varepsilon(A)=\emptyset$.
