Every hereditary graph class is determined by its unique minimal forbidden induced subgraphs

If $G\in\mathcal C$, then no induced subgraph of $G$ lies outside $\mathcal C$, so in particular $G$ contains no member of $\mathcal B(\mathcal C)$.

Suppose $G\notin\mathcal C$. Among vertex sets $W\subseteq V(G)$ for which $G[W]\notin\mathcal C$, choose one of least cardinality; it exists because $W=V(G)$ is available.

Let $\mathcal F$ be any forbidden induced-subgraph basis for $\mathcal C$. Every $J\in\mathcal F$ lies outside $\mathcal C$: otherwise $J\in\mathcal C$ would contain itself as an induced copy of a member of $\mathcal F$, contradicting the defining equivalence for $\mathcal F$.

Every proper induced subgraph of $G[W]$ lies in $\mathcal C$ by minimality of $|W|$. Hence $G[W]\in\mathcal B(\mathcal C)$.

Fix $H\in\mathcal B(\mathcal C)$. Since $H\notin\mathcal C$, it is not $\mathcal F$-free, so some $J\in\mathcal F$ occurs as an induced subgraph of $H$. If that copy were proper, then it would lie in $\mathcal C$ by the minimality of $H$; closure under isomorphism would give $J\in\mathcal C$, contradicting step 1.3. Thus the copy uses all vertices of $H$, and $J\cong H$.

Thus $G$ is not $\mathcal B(\mathcal C)$-free. Together with step 1.1 this proves the equivalence.

Hence every forbidden basis for $\mathcal C$ contains, up to isomorphism, every member of $\mathcal B(\mathcal C)$. Since $\mathcal B(\mathcal C)$ is itself a basis by step 3.1, it is inclusion-minimal, and any inclusion-minimal forbidden basis has no additional members. This proves uniqueness up to isomorphism.