The quotient by a modular partition is isomorphic to the subgraph induced by any set meeting each part exactly once

Define $\phi :\mathcal{P}\to T$ by letting $\phi (M)$ be the unique vertex of $M\cap T$. This is injective, because distinct parts are disjoint and $\phi (M)\in M$; and it is surjective, because every $t\in T$ lies in exactly one part $M$, and then $t\in M\cap T$, so $t=\phi (M)$.

Let $M,N\in \mathcal{P}$ be distinct. The pair $(M,N)$ is complete or anticomplete and not both, so if it is complete then $\phi (M)\phi (N)\in E(G)$, and if it is anticomplete then $\phi (M)\phi (N)\mathrm{\notin }E(G)$.

By the definition of the quotient, $\{M,N\}\in E(G\mathrm{/}\mathcal{P})$ says exactly that $(M,N)$ is complete, so step 1.2 gives $\{M,N\}\in E(G\mathrm{/}\mathcal{P})$ if and only if $\{\phi (M),\phi (N)\}\in E(G)$; and since $\phi (M),\phi (N)\in T$, that is the same as $\{\phi (M),\phi (N)\}\in E(G[T])$.

So $\phi$ is a bijection from $V(G\mathrm{/}\mathcal{P})=\mathcal{P}$ onto $T=V(G[T])$ that preserves and reflects adjacency, hence an isomorphism $G\mathrm{/}\mathcal{P}\to G[T]$.

A set $T$ as in the Statement exists: the parts are nonempty and there are finitely many of them, so selecting one vertex from each is a choice from a finite family of nonempty sets and needs no further principle. Hence the quotient is isomorphic to an induced subgraph of $G$.