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Suppose I have a compact metric space $X$ with an open set $O \subsetneq X$ and a compact set $C \subset O$ given. Is it possible to find another compact set $\tilde{C}$ such that $C \subsetneq \tilde{C} \subset O$?

My idea was to check the distance between the disjoint sets $C$ and $X\setminus O$ and hopefully find a lower bound $\varepsilon > 0$. Then I could expand the set $C$ by $\varepsilon/2$ and take the closure. However, I think the dist-function is not continuous on $C \times X\setminus O$, it is only componentwise contiuous? Is it possible to find such $\tilde{C}$?

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first we need to suppose $C\neq O$ (yes, it could be a clopen). if this is true, take any element $x\in O\setminus C$ and then $C\cup \lbrace x\rbrace$ is close, so compact (compact space)

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    Yes indeed! So then I could write $O = \bigcup C_i$ for all such $C_i$'s. But the index set won't be countable in general. – 2017-02-03
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    My intention was to construct a sequence of compact sets approximating $O$ from below... – 2017-02-03
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    You can use that open sets are $F_{\delta}$, and modify this family to make it monotonous. As the space is compact, al the elements of the family are – 2017-02-03
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    Ah this sounds good, thanks for the hint! :) Didn't know about $F_\delta$ before. – 2017-02-03
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    However, I guess it's called $F_\sigma$, isn't it? – 2017-02-03
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    Yes, every open set $O$ is a union of closed sets $F_n$ (true in all metric spaces), all of which are compact, being closed in $X$. Then $C_n = \bigcup_{i=1}^n F_i$ is an increasing family of compact sets that unions up to $O$. – 2017-02-03
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I'll complete it. Consider the close set $F=X\setminus O$, so compact and the function $f(x)=\frac{\mathrm{d}(x,F)}{1+ \mathrm{d}(x,F)}$, which is continuous and bounded by $1$, and $f(x)=0\iff x\in F$ , so $f:X\rightarrow[0,1]$ and $f^{-1}(0,1]=O$. Then, as $$\bigcup_{n\in\mathbb{N}}[1/n,1]=(0,1]$$ and $[1/n,1]$ are close sets, then if we define $C_i=f^{-1}[1/n,1]$, as f is continuous $C_i$ is close, so compact, and satisfices the asked

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    This is the prove of open are $F_{\delta}$, and actually works here – 2017-02-03