= Solution
A <transitive class> $M$ satisfies $x\in M\Rightarrow x\subseteq M$. Its membership structure therefore does not omit elements of its members. In particular bounded membership statements are absolute between a transitive model and the ambient universe, and its membership relation is the actual well-founded relation, rather than an arbitrary externally ill-founded coding.
For a set $x$, put $x_0=x$ and $x_{n+1}=\bigcup x_n$, and define
$$
\boxed{\operatorname{TC}(x)=\bigcup_{n<\omega}x_n.}
$$
This is a set, contains $x$ as a subset, and is transitive: an element of an element of $x_n$ belongs to $x_{n+1}$. Any transitive set containing $x$ as a subset contains every $x_n$ by induction, so this construction is the least such set. Shifting the union index gives
$$
\boxed{\operatorname{TC}(x)=x\cup\operatorname{TC}(\bigcup x),\qquad
\operatorname{TC}(\{x\})=\{x\}\cup\operatorname{TC}(x).}
$$
These formulas use the specified subset convention for <transitive closure>.
Work in the usual ambient set theory with Choice. If $x$ is a <hereditarily countable set>, then every member of $\operatorname{TC}(\{x\})$ is countable. Iterated countable unions show that this entire closure is countable. Conversely a countable <transitive closure> makes each of its members countable. This equivalent criterion makes the following closure arguments explicit.
If $y\in x\in\mathrm{HC}$, then $\operatorname{TC}(\{y\})\subseteq\operatorname{TC}(\{x\})$, so $y\in\mathrm{HC}$: \b[HC is transitive]. If $x\in\mathrm{HC}$, its union is countable and its descendant sets are descendants of $x$, so $\bigcup x\in\mathrm{HC}$. The actual union therefore witnesses the <Axiom of union> inside $(\mathrm{HC},\in)$.
For <axiom schema of separation>, take any formula with parameters in HC and a set $x\in\mathrm{HC}$. Form the subset of $x$ consisting of the elements that satisfy that formula with all quantifiers relativized to HC. Ambient Separation supplies this set. It is a subset of a <countable set>, and all its descendants are inherited from $x$, so it belongs to HC. No claim that the formula itself is absolute is needed.
The <axioms satisfied by hereditarily countable sets> also include Extensionality, Empty Set, Pairing, Infinity, Foundation and Replacement. Extensionality and Foundation follow from transitivity and ambient well-foundedness. Empty Set and $\omega$ belong to HC, and a pair of HC sets has countable <transitive closure>. For Replacement, the domain is countable; a definable image of its elements is a countable family of HC sets, whose combined transitive closures form a countable union of <countable sets>, so the range again belongs to HC. Ambient Choice additionally gives Choice internally: a well-order of a countable HC set has a hereditarily countable relation and hence lies in HC.
\b[Power Set fails.] Every subset of $\omega$ is hereditarily countable, so an internal power-set witness would have to contain all ambient subsets of $\omega$. That collection is uncountable by <Cantor theorem>, whereas every HC set is countable. Thus HC satisfies the other ZF axioms and, in this ambient setting, Choice, but not the <Axiom of power set>.
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