I am studying with Jech's Set Theory. In the Chapter 6, there is an exercise like this:
Let $B$ be the class of all $x$ that are hereditarily in the class $A$. Show that
(i) $x \in B$ if and only if $\operatorname{TC}(x) \subset A$
(ii) $B$ is the largest transitive class $B \subset A$
Here $\operatorname{TC}$ means transitive closure and $B$ is the unique class satisfying $B = \{x \in A \mid x \subset B\}$. And we assume axiom of regularity(foundation).
I solved (i) for $\implies$ side. Trivially $\operatorname{TC}(\varnothing) = \varnothing \subset A$. Let $x$ be $\in$-minimal such that $x \in B$ and $\operatorname{TC}(x) \not\subset A$. Since $y \in x \in B \implies y \in B$, $\operatorname{TC}(x) = x \cup \bigcup_{y \in x}\operatorname{TC}(y) \subset A$ which is a contradiction. But I can't prove $\impliedby$ side.
For (ii), let $C$ be a transitive class $C \subset A$. Let $x$ be $\in$-minimal such that $x \in C \setminus B$. Then $x \subset B$ so $x \in B$, which is a contradiction.
I think the problem (i) has an error. Assume that (i) is right. Let $A$ be an ordinal. Then $A$ is transitive, so $\operatorname{TC}(A) = B = A$. Then by (i), $\operatorname{TC}(A) \subset A \implies A \in B = A$, which is a contradiction. Am I missing something?
Note that we already quantify over $A$ in the very beginning of the exercise. Namely, we already required that $x\in A$. This renders the situation you describe impossible.
Having said that, I do agree that it would be clearer writing $\operatorname{TC}(\{x\})\subseteq A$.