Let $\Omega \subseteq \mathbb{R}^d$ be bounded, consider the space $X:= \mathcal{B} (\Omega)$ equipped with the metric $d(A,B):=|A \triangle B|$ (where all sets with distance 0 are identified as usual, so we get a metric space $(X,d)$. $d$ is also called Fréchet-Nikodym metric.
It is well-know that X is complete. But is it also compact, i.e. can we show that X is totally bounded?
It was shown here that X contains infinitely many disjoint $\varepsilon$-balls if $\varepsilon$ is small emough: Set of subsets of bounded $\Omega$ with pairwise symmetric difference bounded from below
But this would be no contraction to total boundedness.
It is not compact. Take $\Omega = [0,1]$ in $\mathbb R^1$.
Consider the i.i.d. sequence of "Rademacher" sets \begin{align} A_1 &= \left[0,\frac12\right] \\ A_2 &= \left[0,\frac14\right] \cup \left[\frac12,\frac34\right] \\ A_3 &= \left[0,\frac18\right] \cup \left[\frac14,\frac38\right] \cup \left[\frac12,\frac58\right] \cup \left[\frac34,\frac78\right] \end{align} Then we have $d(A_n,A_m) = \frac12$ for $n \ne m$. The space is not totally bounded.