I want to know if the following theorem is correct.
Let $\mathfrak{B}$ be a Borel $\sigma$-algebra of a Hausdorff topological space $X$. Suppose that $\mu :\mathfrak{B}\to\overline{\mathbb{R}}$ is a locally finite measure. If there's a collection $\{K_n\}_{n\in\mathbb{N}}$ of compacts sets such that $X=\cup_{n\in\mathbb{N}}{K}_n^{\color{red}\circ}$, then for all $B\in\mathfrak{B}$ with $\mu (B)<\infty$ and $\varepsilon\in (0,\infty)$ there're a compact $K$ and an open $O$ such that $K\subseteq B\subseteq U$ and $\mu (O\setminus K)<\varepsilon $.
I will write my proof since it may contain mistakes.
We can use the Corollary 3.4 (which is in the page 74 of the "Measure and Integration" written by John L. Menaldi) to conclude that there're a closed $C$ and an open $O$ such that $C\subseteq B\subseteq O$ and $\mu (O\setminus C)<\varepsilon $.
Define $\tilde K_n:=C\cap (\cup_{i=1}^nK_i)$ for all $n\in\mathbb{N}$. Then $\{\tilde K_n\}_{n\in\mathbb{N}}$ is a collection of compact sets such that $\tilde K_n\nearrow C$ with implies that $O\setminus \tilde K_n\searrow O\setminus C\,\,\color{red}{(1) }$.
Since $C\subseteq B\subseteq O$, then $O=(O\setminus B)\sqcup B\subseteq (O\setminus C)\cup B$ which implies that $\mu (O)\leq \mu (O\setminus C)+\mu (B)<\infty$ since $\mu (O\setminus C)<\varepsilon $ and, by hypothesis, $\mu (B)<\infty$.
Therefore $\mu (O\setminus \tilde K_1)<\infty$. With this inequality and $(1)$ we can conclude that $\lim_{n\to\infty}\mu _X(O\setminus \tilde K_n)=\mu _X(O\setminus C)<\varepsilon $ and, consequently, that there's $N\in\mathbb{N}$ such that $\mu _X(O\setminus \tilde K_N)<\varepsilon $. We conclude the proof since $\tilde K_N$ is a compact satisfying $\tilde K_N\subseteq C\subseteq B\subseteq A$.
I think that the proof above contains an error since with the above theorem we can conclude that $\mu$ is inner regular.
Thank you for your attention!