Wrong Proof of Secure PRF
The problem is in the last equation:
$$
Pr[\mathcal{A}^{\{U\}}(1^{\lambda})=1]=Pr[\mathcal{A'}^{\{U\}}(1^{\lambda})=1]
$$

It does not hold because $\mathcal{A}^{\{U\}}$ is the result of $\mathcal{A}'$ using $U$ instead of $F_k$. This is actually equivalent to $F'$ and will be distinguished by $\mathcal{A}'$, so we get
$$
Pr[\mathcal{A}^{\{U\}}(1^{\lambda})=1]=Pr[\mathcal{A'}^{\{F'(U)\}}(1^{\lambda})=1]\neq Pr[\mathcal{A'}^{\{U\}}(1^{\lambda})=1]
$$