The Halting problem proof is wrong?
First, let us see what the halting proof attempts to prove:


  There is no program $H$ that, on input $(x,y)$, always halts, and returns whether the program encoded by $x$ halts when run on the input $y$.


We call the function which $H$ is supposed to compute the halting predicate.

The program you are suggesting, which consists of simulating a run of program $x$ on input $y$, definitely doesn't always halt. What we are attempting to show is that there is no other devious way to computing the halting predicate in a way which does always halt.

The proof you are describing is a proof by contradiction. We assume that there is a program $H$ that always halts and computes the halting predicate correctly. Under this assumption, we reach a contradiction. Hence the conclusion is that no such program exists.


  How does $H$ know whether $x$ halts on $x$?


The assumption that the proof by contradiction is making is that there does exist an $H$ which computes the halting predicate. What the proof shows is that $H$ actually doesn't compute the halting predicate correctly. So $H$ cannot tell whether $x$ halts on $x$, and that is exactly what the proof shows.


  The answer is $H$ must simulate $x$ on $x$.


This is not true. There are similar predicates which can be computed. For example, we can decide whether a program halts on all inputs within 100 steps. We can do so even though there are infinitely many potential inputs, and so the naive solution, which is to run the program on all inputs for 100 steps, can be improved.

What the proof shows is that there is no computable way to decide whether a given program $x$ halts on a given input $y$. It doesn't show that $H$ must simulate $x$ on $y$; indeed, $H$ could store the answers for selected pairs $(x,y)$ in a table, and for these pairs it could just return the correct answer by doing a table lookup. This approach can only handle finitely many pairs $(x,y)$, and the proof shows that no other computable approach works.