Let $A$ be a connected subset in $ \Bbb R^n$ and $ \epsilon >0$ then the $ \epsilon$ -neighbourhood of $A$ is path-connected. Let $A$ be a connected subset in $ \Bbb R^n$ and $ \epsilon >0$ then the  $ \epsilon$ -neighbourhood of $A$ defined by $U_\epsilon(A) := \{x \in \Bbb R^n : d_A(x) < e\}$ is path-connected.
If $A$ is a circular space in $ \Bbb R^n$ then I can visualize the geometry very easily. For any two points in $U_\epsilon(A)$ we draw tangents to the circle, if the two tangents meet then we are done. If not then we find the two points where they meet the $ \epsilon$ -neighbourhood and draw a line between them.
But how to do it in general?
 A: An idea: For $x,y \in U_\epsilon(A),$ define $x\sim y$ to mean there is a path from $x$ to $y$ in $U_\epsilon(A).$ This is an equivalence relation on $U_\epsilon(A).$ The equivalence classes induced by $\sim$ should be open (and of course pairwise disjoint). If there is more than one such equivalence class, we should should be able to get a separation of $A,$ contradiction.
A: suppose this is not path-connected. fix a point $$x\in U_\varepsilon (A)$$consider $\ U=\{{y \in  U_\varepsilon(A)|}$ there is a path between x and y}$\\$
$U_\varepsilon(A)$ is open, so you can prove that the set $U$ is also open.
 we claim that $U^c$ is also open. proof: if there is no path between x and y so there is no path between x and each neighborhood of y . $\Rightarrow U^c$ is open $\Rightarrow U $is both open and close, so$ U=\varnothing$ or $U=U_\varepsilon(A)$. but we know that $U$ is not empty since there is always a path from the point x into itself! so $$U=U_\varepsilon(A)$$ and all of this neighborhood is path-connected.
