open covering for compact set exist $\delta$ neighborhood covering Let $K$ be a compact set in metric space $\Bbb{R}^n$.Prove for an open covering $\mathcal{O}_\alpha$ exist finite open covering such that those set not only covers $K$ but also $\delta$ neiborhood of $K$.that is exist some $\delta>0$ such that subcovering $\mathcal{O}_i$ $i = 1,...,n$ convers $K_\delta = \{x:\text{dist}(K,x)<\delta\}$.
I try to solve it type of arguement in  Lebesgue number lemma  as follows:
Let finite covering be $\mathcal{O}_i$ ,first we can refine the covering to be the compact included convering(as my previous post shows here).Then exist a compact set $\bigcup \mathcal{O}_i\subset C$.then we know $C\setminus\mathcal{O}_i = V_i$ is compact set.then we know continuous function $f:K\to \Bbb{R}$ which is
$$x\mapsto 1/n \sum d(x,V_i)$$
is always positive,hence exist minimum value $\delta$ which can be shown any $x\in K_\delta$,exist some $\mathcal{O}_i$ covers it using the reation that average value is always less than maximum value.
Is my proof correct.
 A: I think there might be problems in your last line of reasoning.
First, let me settle some notation. If $O_1,\dotsc, O_n$ is your original finite cover, then let $U_i$ be the compactly included refinement. We have
$K\subset \bigcup_i U_i\subset C\subset \bigcup_i O_i$ where $C$ is the union of the closures of the $U_i$. Also, we put $V_i =C\setminus U_i$, which are each compact.
I think your desired argument goes something like this: if $z\in K_\delta$ we have $d(z,x)<\delta$ for some $x\in K$. On the other hand, $d(x,V_i)\geq\delta$ for some $i$ (since otherwise $f(x)<\delta$, contradicting that $\delta$ was the minimum). Now you want to say $z\in U_i$, since otherwise $z\in V_i$ and we get $d(x,V_i)\leq d(x,z)<\delta$, a contradiction. This only works though if we assume $z\in C$ to begin with. The most we can say is that $z\in (C\setminus U_i)^c = C^c\cup U_i$.
Here's one possible fix: define the $V_i$ to be $U_i^c$ instead of $C\setminus U_i$. After all, we don't need $V_i$ to be compact for $x\mapsto d(x,V_i)$ to be continuous. And we still know $f(x)>0$ for $x\in K$ since otherwise $x\in V_i$ for each $i$, which would in turn mean $x\notin \bigcup U_i\supset K$.
Honestly though, this result can be proven with much less machinery. Let ${\cal O}= \{O_\alpha: \alpha\in A\}$ be an open cover of $K$. For each $x\in K$, there is some element of this collection $O_{\alpha_x} = O_x$ which contains $x$. Also, by openness, there is some $\delta_x>0$ such that $B(x,2\delta_x)\subset O_x$.
Now it's not hard to see that the collection $\{B(x,\delta_x): x\in K\}$ is an open cover of $K$, and hence there are finitely many points $x_1,\dotsc, x_n$ such that
$$K\subset B(x_1,\delta_1)\cup\dotsm\cup B(x_n,\delta_n).$$
Yet now we've given ourselves some wiggle room. Denoting $O_k = O_{x_k}$, we actually have
$$K\subset B(x_1,\delta_1)\cup\dotsm\cup B(x_n,\delta_n)\subset B(x_1,2\delta_1)\cup\dotsm\cup B(x_n,2\delta_n)\subset O_1\cup\dotsm\cup O_n.$$
From here, put $\delta = \min (\delta_1,\dotsc, \delta_n)$. You can easily see by triangle inequality that each point of $K_\delta$ belongs to one of the $B(x_k,2\delta_k)$ balls, and in particular is covered by $O_k$.
