Completeness problem of TM
Assume $H$ does not halt on $w$. To prove $L(A) = \Sigma^∗$ we need to show that $A$ accepts every $x$. Indeed, whatever $x$ is, we simulate the execution of $H$ on $w$ for only $|x|$ steps (a finite, known amount of steps). I.e., we do not simply run that program on its own, we keep an "alarm clock" counting the steps $H$ as it is running, and as soon as it reaches $|x|$ the alarm rings and we stop the simulation. In this way, no matter what $H$ does on $w$, the simulation will always terminate: either because $H$ halts before $|x|$ steps, or because the alarm rings and we abort the simulation.

Now, since $H$ does not halt on $w$ by hypothesis, $H$ will always make the alarm ring. In such case, $A$ accepts $x$ by construction, hence $A$ indeed accepts all inputs $x$.

For the other direction, assume that instead $H$ does halt on $w$. Let's write $k$ for the number of steps that make $H$ halt on $w$. We now need to prove that $L(A) \neq \Sigma^∗$. Take $x = 1^k$: we prove that $A$ rejects it. By definition, $A$ simulates $H$ on $w$ for $|x|=k$ steps, and that's just enough time to make $H$ halt. So, by construction $A$ rejects.