First consider $A\subset \mathbb{R}^n$ to be a bounded open set and designate the open ball centered at $x$ with radius $r$ by $B_r(x)$

Construct $C_n=\bigcup_{x \in \partial A} B_{\frac{1}{n}}(x)$ , where $\partial A$ is the boundary of $A$ and define $K_n = A \backslash C_n$ .

• $K_n$ is compact.

  It is bounded since $K_n \subset A$ and $A$ is by assumption bounded. $K_n$ is also closed. To see this consider $x \in \partial K_n$ but $x \notin K_n$ . Then there exists $y \in \partial A$ and $0<r<\frac{1}{n}$ such that $x \in B_r(y)$ . But $B_r(y) \cap K_n = \{\}$ because $B_{\frac{1}{n}}(y) \cap K_n = \{\}$ and $0<r<\frac{1}{n} \implies B_r(y) \subset B_{\frac{1}{n}}(y)$ . This implies that $x \notin \partial K_n$ and we have a contradiction. $K_n$ is therefore closed.

• $K_{n} \subset \inter K_{n+1}$

  Suppose $x \in K_n$ and $x \notin \inter K_{n+1}$ . This means that for all $y \in \partial A$ , $x \in \overline{B_{\frac{1}{n+1}}(y)} \vee x \in \mathbb{R}^n \backslash A$ . Since $x \in K_n \implies x \in A$ we must have $x \in \overline{B_{\frac{1}{n+1}}(y)}$ . But $x \in \overline{B_{\frac{1}{n+1}}(y)} \subset B_{\frac{1}{n}}(y) \implies x \notin K_n$ and we have a contradiction.

• $\bigcup_{n=1}^{\infty} K_n = A$

  Suppose $x \in A$ , since $A$ is open there must exist $r>0$ such that $B_r(x) \subset A$ . Considering $n$ such that $\frac{1}{n}<r$ we have that $x \notin B_{\frac{1}{n}}(y)$ for all $y \in \partial A$ and thus $x \in K_n$ .

Finally if $A$ is not bounded consider $A_k = A \cap B_{k}(0)$ and define $K_n = \bigcup_{k=1}^n K_{k,n}$ where $K_{k,n}$ is the set resulting from the previous construction on the bounded set $A_k$ .

• $K_n$ will be compact because it is the finite union of compact sets.
• $K_{n} \subset \inter K_{n+1}$ because $K_{k,n} \subset \inter K_{k,n+1}$ and $\inter(A \cup B) \subset \inter A \cup \inter B$
• $\bigcup_{n=1}^{\infty} K_n = A$

  First find $k$ such that $x \in A_k$ . This will always be possible since all it requires is that $k>|x|$ . Finally since $n > k \implies K_{k,n} \subset K_{n,n}$ by construction the argument for the bounded case is directly applicable.