The Yoneda lemma

We first check $\Phi(\Psi(x))=x$ for every $x\in F(A)$. Unwind the definitions: $\Psi(x)_A(1_A)=F(1_A)(x)$ by the formula for $\Psi$, specialized to $X=A$, $f=1_A$. Since $F$ is a functor, it sends the identity morphism $1_A$ to the identity function on $F(A)$, so $F(1_A)(x)=x$. Therefore $\Phi(\Psi(x))=\Psi(x)_A(1_A)=x$, exactly as required — this direction uses nothing but the functor identity law.

Now we check the harder direction, $\Psi(\Phi(\eta))=\eta$ for every natural transformation $\eta:h_A\Rightarrow F$. Both sides are natural transformations $h_A\Rightarrow F$, so it suffices to show their components agree at every object $X$ and every element $f\in h_A(X)=\mathrm{Hom}(X,A)$.

Unwind $\Psi(\Phi(\eta))_X(f)$ using the formula for $\Psi$ with $x:=\Phi(\eta)=\eta_A(1_A)$: this equals $F(f)(\eta_A(1_A))$.

Now invoke naturality of $\eta$ itself, in the form $\eta_Y(f\circ g)=F(g)(\eta_X(f))$ with $g:=f:X\to A$ and the other variable set to $A$: taking $f:=1_A\in h_A(A)$ in that naturality square gives exactly $\eta_X(1_A\circ f)=F(f)(\eta_A(1_A))$, i.e. $\eta_X(f)=F(f)(\eta_A(1_A))$, since $1_A\circ f=f$.

Combining the last two displays: $\Psi(\Phi(\eta))_X(f)=F(f)(\eta_A(1_A))=\eta_X(f)$. Since $X$ and $f$ were arbitrary, the two natural transformations agree everywhere, so $\Psi(\Phi(\eta))=\eta$. Together with the first paragraph, $\Phi$ and $\Psi$ are mutually inverse bijections.