Wrong Proof of Secure PRF Disclaimer, this is not homework; I found the question statement here: Security of this PRF

To restate the question:


  Given $F$ a secure PRF with input size $\lambda$. Define $F'$ as
  $F'(k,x\mathbin\|x') = F(k, 0\mathbin\|x)\oplus F(k, 1\mathbin\|x')$
  with $x$ and $x'$ of $\lambda-1$ bits.
  Is $F'$ a secure PRF?


The given answer is no and it's clear that the PRF $F'$ is not secure. But I tried to provide a false proof that $F'$ is secure (but it's not) and cannot see what step the proof goes wrong. 

First, to restate the counterexample:


  More specifically, for sake of simplicity let $F'_k$ be a keyed PRF
  $F'$ with key $k$ from the family of unkeyed PRFs $\{F' : \{0,
 1\}^{2(\lambda - 1)} \times \{0, 1\}^{2(\lambda - 1)} \to \{0,
 1\}^{\lambda - 1} \} $ and consider the following  $\mathsf{PPT}$
  adversary $\mathcal{A}$. $\mathcal{A}$ asks for the evaluation of the
  oracle $\mathcal{O}$ on $x_0 \mathbin\| x_1$, $x_0 \mathbin\| x_2$, $x_1 \mathbin\| x_2$, and
  $x_1 \mathbin\| x_1$. Then $\mathcal{A}$ computes the XOR of the first three
  oracle outputs and checks if the XOR is equal to the fourth oracle
  output. Observe that if $\mathcal{A}$ is given oracle access to
  $F'_k$, then $\mathcal{A}$ receives \begin{align*} F'_k (x_0 \mathbin\| x_1)
 &amp;= F_k (0 \mathbin\| x_0) \oplus F_k (1 \mathbin\| x_1)\\ F'_k (x_0 \mathbin\| x_2) &amp;= F_k (0
 \mathbin\| x_0) \oplus F_k (1 \mathbin\| x_2)\\ F'_k (x_1 \mathbin\| x_2) &amp;= F_k (0 \mathbin\| x_1)
 \oplus F_k (1 \mathbin\| x_2)\\ F'_k (x_1 \mathbin\| x_1) &amp;= F_k (0 \mathbin\| x_1) \oplus F_k
 (1 \mathbin\| x_1) \end{align*} 
  and the XOR of the first three is exactly
  equal to the fourth and $\mathcal{A}$ guesses correctly with
  probability 1 given this case of the oracle. In the random case,
  $\mathcal{O}$ is equal to some truly random function $U$ and receives
  $U(x_0 \mathbin\| x_1) \oplus U (x_0 \mathbin\| x_2) \oplus U (x_1 \mathbin\| x_2)$ is
  unlikely to be equal to $U (x_1 \mathbin\| x_1)$, namely the probability is
  negligible. Hence $\mathcal{A}$ is a distinguisher.


Here is my "false proof" in an attempt at proving the statement "$F'$ is a secure PRF" and I'd really appreciate clarification on what step is incorrect.

Suppose $F_k'$ is not a PRF. Then there exists a $\mathsf{PPT}$ distinguisher $\mathcal{A}'$ and a constant $c$ and natural number $n \in \mathbb{N}$ such that $\mathcal{A}'$ distinguishes $F_k'$ from a truly random function with distinguishing advantage $\geq \frac{1}{n^c}$.
We now construct a distinguisher $\mathcal{A}$ (an oracle machine with access to $\mathcal{O}$ which is either from PRF family $\{F\}$ or truly random function $\{U\}$) for $F$ as follows.
$\mathcal{A}$ runs $\mathcal{A}'$. $\mathcal{A}'$ asks for queries of the form $x_0 \mathbin\| x_1$ as described above and $\mathcal{A}$ then queries the oracle $\mathcal{O}$ with $0 \mathbin\| x_0$ and $1 \mathbin\| x_1$, computes the XOR $\mathcal{O}(0 \mathbin\| x_0) \oplus \mathcal{O}(1 \mathbin\| x_1)$, and sends that to $\mathcal{A}'$.
$\mathcal{A}'$ will query polynomially many times in $\mathcal{A}'$'s input length so $\mathcal{A}$ will have to make twice as many, namely $O((2(\lambda-1))^l)$ queries for some constant $l$, which is still polynomial in $\lambda$. 
Finally, $\mathcal{A}$ will output whatever $\mathcal{A}'$ outputs.
Note then $\Pr[\mathcal{A}^{\{F\}}(1^{\lambda}) = 1] = \Pr[\mathcal{A}'^{\{F'\}}(1^{2(\lambda -1)}) = 1]$ by construction since $\mathcal{A}$ effectively simulates oracle access to $\{F'\}$ when $\mathcal{O} = F_k$ for some $k$.
If $\mathcal{O} = \{U\}$ for truly random functions $U$, then $\Pr[\mathcal{A}^{\{U\}}(1^{\lambda}) = 1] = \Pr[\mathcal{A}'^{\{U\}}(1^{2(\lambda -1)}) = 1]$ because $U(0 \mathbin\| x_0) \oplus U(1 \mathbin\| x_1)$ is still truly random. 
Then the distinguishing advantage of the new adversary $\mathcal{A}$ is also non-negligible and hence $F$ is not a secure PRF.

Again, where did I go wrong in this "false proof"?
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
$$
Pr[\mathcal{A}^{\{U\}}(1^{\lambda})=1]=Pr[\mathcal{A'}^{\{F'(U)\}}(1^{\lambda})=1]\neq Pr[\mathcal{A'}^{\{U\}}(1^{\lambda})=1]
$$