For every class $C$, find the unique class of "hereditarily C sets". My textbook claims (without proof, which I'm trying to see for myself) that for every class $C$ there exists a unique class HC such that
$$\forall x (x\in HC \iff (x \in C \wedge \forall y \in x, y\in HC)$$
And the text calls HC the class of "hereditarily C sets". 
This is in $ZF^{--}_F$.

First, is this a typo, is it mean to say "hereditary C sets"? 
According to wikipedia (https://en.wikipedia.org/wiki/Hereditary_set) in pure set theory all sets are hereditary because the only property we describe as hereditary is that of being a set, and our only objects of discourse are sets. Is the answer here obvious then by taking HC to be the universe class $V=\{x| x=x\}$?
 A: EDIT Thanks to Eric Wofsey for correcting some errors I made in the original answer.
As discussed in the comments, we can use the transitive closure to make the definition explicit, defining HC as $$ x\in HC \iff x\in C\land (\forall y \in \operatorname{trcl}(x))(y\in C).$$ The usual definition of the transitive closure is by a recursion of length $\omega$
$$ G(x,0) = x\\G(x,n+1) = \cup G(x,n)\\ \operatorname{trcl}(x) = \bigcup_{n\in\omega} G(x,n)$$ 
The fact that we don't have infinity causes an issue here. The recursion theorem still works and produces the class function $G,$ but the union in the last line does not necessarily produce a set. 
But as far as I can tell, we only need $\operatorname{trcl}(x)$ as a class to make this work, so the definition can be carried out in the theory your book calls ZF$^{--}$ (i.e. foundation isn't required). 
(Note in my original answer I thought that foundation could also be used to resolve this via $\in$-recursion. It turns out that was wrong and that foundation has some subtleties in the absence of infinity. See the comments.)
