Prove that metric spaces form a proper class (MK axiomatics) I am trying to solve this exercise, and I really haven't done nothing similar yet. So I will give my idea but, other than being incomplete, it may be far from being correct.
I would try something like:
Consider the class of compact metric spaces. If they formed a set $K$, rather than a proper class, then I could endow $K$ with a metric like the Gromov-Hausdorff one. Assuming this turns $K$ into a compact metric space (and I have no clue about this) then we reach the paradox $K\in K$. The collection of all metric spaces contains that of compact metric spaces and a fortiori it will not be a set. 
I don't hope this argument to be correct, but is it at least plausible? Can we save it somehow or carry out a similar one directly on the collection of all metric spaces? ( The Gromov-Hausdorff distance is the only distance between metric spaces I know) 
Otherwise, how to tackle the problem of proving that the collection of sets with some structure form ( or does not form) a proper class?
 A: The class of all finite sets is a proper class. And every finite set has a unique metric topology (the discrete one) making it a compact metric space. So "all metric spaces" contains a proper class and thus is a proper class itself.
A: In general, any class of objects which are "(nonempty) sets with structure," where the underlying set of such an object isn't important, is a proper class. This is because we can always just tweak the underlying set of a member of the class and stay in the class.
For example:


*

*The class of groups is a proper class, since $(i)$ every cardinality-$42$ set is the underlying set of some group and $(ii)$ the class of cardinality-$42$ sets is a proper class.

*The class of metric spaces is a proper class, since $(i)$ every cardinality-$\aleph_{17}$ set is the underlying set of some metric space (use the discrete metric) and the class of cardinality-$\aleph_{17}$ sets is a proper class.

*In the two examples above, we could use any cardinality $(>0)$ we want. Sometimes we have to be a bit more careful: for example, every separable metric space has cardinality $\le 2^{\aleph_0}$, so we need to pick a cardinality $\le 2^{\aleph_0}$ to use. It's probably best here (and, frankly, whenever possible) to just work with $1$: $(i)$ every $1$-element set is the underlying set of some metric space and $(ii)$ the class of $1$-element sets is a proper class.

A bit more abstractly, here's what's going on. If $\mathcal{C}$ is a class of the above type, pick some (nonempty) $\mathfrak{X}\in\mathcal{C}$. Let $X$ be the "underlying set" of $\mathfrak{X}$; then any set $Y$ with $\vert X\vert=\vert Y\vert$ is the underlying set of some $\mathfrak{Y}\in\mathcal{C}$, gotten by "porting over" the structure of $\mathfrak{X}$ onto $Y$ via any fixed bijection $X\cong Y$.
Since for any (nonzero) cardinal $\kappa$ the class of cardinality-$\kappa$ sets is a proper class, this shows that $\mathcal{C}$ is a proper class.
