Let $K⊂\mathbb{R}^d,d∈\mathbb{N}$ be compact and $ε>0$. Show that $∪_{x∈K},K_ε(x)$ is compact. [Updated Question] Let $K⊂\mathbb{R}^d,d∈\mathbb{N}$ be compact and $ε>0$. Show that $∪_{x∈K},K_ε(x)$ is compact.
$K_ε(x)$ is a $ε$-ball in the form of $K_ε(x) = \{x ∈ \mathbb{R}^d
: ∥x-y∥ ≤ ε\} = \overline{B_ε(x)}.$
Does anyone have a proof on how to show the statement in the title?
 A: Consider the continuous  function $F: \mathbb R^d \times \mathbb R^d \to  \mathbb  R^d$ given by $F(x,y) = x+y$.
The closed ball $\overline{B_ε(x)}$ is the set $ x +   \overline{B_ε(0)} $ for $B_ε(0)$ the closed ball centred at the origin.
It follows the set   $\bigcup_{x∈K}K_ε(x)=\bigcup_{x∈K}\overline{B_ε(x)}$ can be written $F\big (K \times \overline{B_ε(0)}\big)$.
But this is the image of a compact set under a continuous function. Hence it is compact.
A: Let $(u_n)$ a sequence of $\bigcup_{x\in K}K_\varepsilon (x)$. We want to prove that $(u_n)$ has a convergent subsequence. Let $x_n\in K$ s.t. $u_n\in K_\varepsilon (x_n)$ for all $n$. Since $(x_n)\in K^{\mathbb N}$ there is a subsequence (still denoted $(x_n)$) that converges. Denote $x\in K$ its limit. Let $N$ big enough so that $|x_{n}-x|<\varepsilon  $ for all $n\geq N$. In particular, $$|u_{n}-x|\leq |u_{n}-x_{n}|+|x_{n}-x|\leq 2\varepsilon $$ for all $n\geq N$, and thus $(u_{n})$ is bounded. Therefore, $(u_n)$ has a convergence subsequence. Denotes $u$ its limit.
For all $n\in \mathbb N$, there is $m_n>n$ s.t. $|u_{m_n}-u|<\frac{1}{n}$ and $|x_{m_n}-x|<\frac{1}{n}$. In particular $$|u-x|\leq |u-u_{m_n}|+|u_{m_n}-x_{m_n}|+|x_{m_n}-x|\leq \frac{1}{n}+\varepsilon +\frac{1}{n}\underset{n\to \infty }{\longrightarrow}\varepsilon .$$ Therefore, $u\in K_\varepsilon (x)$. This prove that $\bigcup_{x\in K}K_\varepsilon (x)$ is sequentially compact, and thus compact.
