Suppose that an open set $U$ of $\mathbb{R}^n$ and a compact subset $K$ of $\mathbb{R}^n$ satisfy $K \subset U$. Show that there exists an open set $V$ of $\mathbb{R}^n$ which satisfies all of the following conditions: $$K \subset V, \text{Cl}(V) \subset U, \text{Cl}(V) \ \text{is} \ \text{compact}.$$ $\text{Cl}(V)$ denotes the closure of $V$ in $\mathbb{R}^n$.
This is a topological space problem. I know intuitively that it is correct, but I cannot figure out how to describe it mathematically rigorously. Here is what I have come up with:
Since $\text{Bd}(K)\cup\text{Bd}(U)=\emptyset$, we can take the $\epsilon$ such that $$\epsilon=\inf\{d(u, k)\mid u\in \text{Bd(U)}, k\in \text{Bd(K)}\}$$ and let $$V=K\cup\{B(k, \frac{\epsilon}{2})\mid k\in\text{Bd(K)}\}$$ Then this $V$ would satisfy the given conditions, which I failed to prove.
I would appreciate any help you can give me. Thank you in advance.
The existence of $V$ open with $K \subseteq V \subseteq \overline{V} \subseteq U$ easily follows from $\Bbb R^n$ being Hausdorff and locally compact. In a general space already.
No need for a metric or Euclidean specific argument.