The Yoneda embedding is fully faithful

Apply the Yoneda lemma with the target functor set to $F:=h_B$. The lemma gives a bijection $\mathrm{Nat}(h_A,h_B)\cong h_B(A)$, and unwinding the definition $h_B(A)=\mathrm{Hom}_{\mathcal{C}}(A,B)$ gives exactly $\mathrm{Nat}(h_A,h_B)\cong\mathrm{Hom}_{\mathcal{C}}(A,B)$.

It remains to identify what morphism of $\mathcal{C}$ a natural transformation $\eta:h_A\Rightarrow h_B$ corresponds to under this bijection, and to check it is postcomposition. By the formula for $\Phi$, $\eta$ corresponds to the element $\Phi(\eta)=\eta_A(1_A)\in\mathrm{Hom}(A,B)$; call this morphism $f$.

Now use $\Psi(\Phi(\eta))=\eta$ (the theorem above, applied with $F=h_B$) to recover every component of $\eta$ from $f$: for any object $X$ and any $g\in h_A(X)=\mathrm{Hom}(X,A)$, $\Psi(f)_X(g)=h_B(g)(f)$. Unwinding the contravariant action of $h_B$ on morphisms — precomposition — gives $h_B(g)(f)=f\circ g$. So $\eta_X(g)=f\circ g$: every component of $\eta$ is literally postcomposition with $f$.

Because $\Phi$ and $\Psi$ are mutually inverse bijections (proved above) and postcomposition with $f$ is exactly $\Psi(f)$, the assignment sending $f$ to postcomposition with $f$ is itself a bijection $\mathrm{Hom}(A,B)\to\mathrm{Nat}(h_A,h_B)$ — which is precisely the statement that $y$ is fully faithful.