$A \subseteq \mathbb{R}^n$ closed and connected. Prove $\{x \in \mathbb{R}^n \mid d(x, A) \le \varepsilon\}$ is path-connected I've encountered the following question:
Say $A \subseteq \mathbb{R}^n$ is a closed and connected set. Prove $\{x \in \mathbb{R}^n \mid d(x, A) \le \varepsilon\}$ is path-connected.
I'm not really sure how to approach this question. It appears under the section of Compactness.
Any help will be appreciated!
 A: Hint: For any connected (nonvoid) subset $A$ of $\mathbb{R}^n$ the set $\{x\ | \ d(x,A) < \epsilon\}$ is connected, since it is the union $\cup_{a\in A} B(a,\epsilon)$, every ball $B(a,\epsilon)$ is connected, and also $A$ is connected ( think of a fishbone). Moreover, $\{x\ | \ d(x,A) < \epsilon\}$ is open, therefore, $\{x\ | \ d(x,A) < \epsilon\}$ is $path\ connected$. 
Connect any point $y$ in  $\{x\ | \ d(x,A) \le \epsilon\}$ with a closest point in $a \in \bar A$, the half open segment $(y,a]$ will lie in $\{x\ | \ d(x,A) < \epsilon\}$ (this is the idea of @Omnomnomnom ). 
A: Hint: let $S = \{x \in \mathbb{R}^n | d(x, A) \le \varepsilon\}$.  Consider any $x_1,x_2 \in S$.  There exist points $y_1,y_2 \in A$ such that $d(x_i,y_i) \leq \epsilon$ for $i = 1,2$.


*

*Show that there is a path connecting $x_i$ and $y_i$ for $i = 1,2$.

*Show that there is a path connecting $y_1$ to $y_2$

*Conclude that there is a path connecting $x_1$ to $x_2$


For the first part, we can consider the path
$$
x(t) = (1-t)x_i + t y_i, \quad t \in [0,1]
$$
