Let $(X,d)$ be a separable metric space and $A\subseteq X$ be compact.  

Since every finite set is doubling then, the collection $\mathcal{A}$ of doubling subsets of $A$ cannot be empty.  My *initial* question was, how large can the following be
$$
\sup_{\tilde{A}\in \mathcal{A}}\,d(A,\tilde{A})
?
$$
*Where $d(A,\tilde{A})$ denotes the [Hausdorff distance][1] between $A$ and $\tilde{A}$.*

However, I have a feeling that the above quantity is $0$ if $A$ is not doubling by the separability of $X$.  

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So then, let me refine my question: For any $\tilde{A}\subseteq A$ and any $r>0$ let $N(\tilde{A},r)$ denote the smallest number of balls of radius $r$ required to cover $\tilde{A}$?

For any $C,d>0$ how large can the following be: 
$$
\begin{aligned}
\sup_{\tilde{A}\in \mathcal{A}}& \,d(A,\tilde{A})\\
\mbox{s.t.}\, & N(\tilde{A},r) \leq C(|A|/r)^d \mbox{for all $0<r\leq |\tilde{A}|$}
\end{aligned}
$$


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*Alternatively, is there a way to quantify how far a metric space is from being doubling?*


  [1]: https://en.wikipedia.org/wiki/Hausdorff_distance