Imagine that the malicious adversary in the real world uses $Y = 0$ as its input, therefore he computes the output as ($X$ && $y=0$) = $0$. We can conclude that the real view in the real world is as:
Real view: $X, Y = 0$ and $output = 0$
Now, imagine the simulator wants to simulate the real world’s view in the ideal world. In order to simulate $x$ (honest party’s input), the simulator must send $1$ to the trusted party and then receives $x$ from it. Thus the simulator succeeded to simulate $X$, but now the simulator computes the output as $X$ && $(Y = 1)$ = $X$, which is not the equal of the output in the real view $X$ && $0$ = $0$. Thus the simulator cannot simulate the output in the ideal model.
Definition 2.3.1 in the malicious model
The definition says that for every adversary A for the real model must exist a simulator S for the ideal model. Considering this definition, I could find an adversary (who sends $y = 0$ in the real world) that there is no any simulator for it, so the protocol is not secure in the malicious model. In the Lindell’s book-page 27 (the below proof), it is said that this protocol is secure!!! I am so confused. (I found a scenario where the protocol is not secure).
Proof of the protocol