Let for an open connected subset $U\subset \mathbb R^n$ and a number $\varepsilon >0$: $$ U_\varepsilon=\{x\in U: dist (x, \partial U)> \varepsilon \}. $$ Then $U_\varepsilon$ is open but in general not connected. But I suppose that for sufficiently small $\varepsilon >0$ also $U_\varepsilon$ is connected. How to prove its?
Edit. Let's assume that additionally $U$ is bounded. Is then $U_\varepsilon$ connected?
No.
Let $$U=\bigcup_{n=1}^\infty B\left((2n+2^{-n},0),1\right)$$ Then $U$ is connected, but for $0<\epsilon<1$, we find that for all $n$, the ball $B((2n+2^{-n}),1-\epsilon)$ is contained in $U_\epsilon$, but for sufficiently large $n$ there is a gap between the $n$th and the $(n+1)$st ball. Indeed, by Pythagoras the width "neck" between consecutive balls is $$ \sqrt{2^2-\left(2+2^{-(n+1)}-2^{-n}\right)^2}=\sqrt{2^{-n}-2^{-2(n+1)}}\approx 2^{-n/2}\to 0$$ and hence $<2\epsilon$ for $n$ big enough.
The problem has changed to add the assumption that $U$ be bounded. Consider (make a sketch!) $$\bigcup_{n=1}^\infty B((2^{-n},0),2^{-n-2})\cup\{\,(x,y)\mid |y|<x^2<1 \}$$ Now for given (small) $\epsilon>0$, find $n$ such that $2^{-n-2}>2\epsilon\ge 2^{-(n+1)-2}$, i.e. the $n$th ball is the last one contributing to $U_\epsilon$. Let $x_0=\sqrt2\cdot 2^{-n}$. Then $x_0>2^{-n}+2^{-n-2}=\frac542^{-n}$ and $x_0<2^{-(n-1)}-2^{-(n-1)-2}=\frac32 2^{-n}$, i.e. $x_0$ is between balls $n$ and $n-1$. At this position, the "bridge" is only $2x_0^2=2^{1-2n}$ wide and this is $<2\epsilon$ as soon as $n\ge 4$.