How a reduction can help up solve a problem?
This is a natural trap that almost everyone falls into when encountering this style of reduction for the first time - you have your logic inverted, but in an understandable way.

To start with, note a couple of things:


If $N$ accepts at least one string, it accepts every string, so $L(N)$ is either $\emptyset$ or $\Sigma^{\ast}$, nothing in between.
$\langle N \rangle$ is an encoding of a Turing Machine, and what we want to know is whether we can decide if $\langle N \rangle$ is in $L$ or not.


So the argument is by contradiction. Imagine that we did have a TM $X$ that decided $L$, then we could give it $\langle N \rangle$ and, as it decides $L$, it would say, correctly, $\mathrm{Yes}$ or $\mathrm{No}$.

But if it says $\mathrm{Yes}$, then we know that $M$ halts on $x$, because that's how we built $N$, and similarly if it answers $\mathrm{No}$, $M$ doesn't halt on $x$.

So having $X$ (i.e. being able to decide $L$) allows us to decide the Halting Problem, and hence we have a contradiction.

Note that we're not trying to 'run' $N$, we're asking if there can possibly be an $X$ that looks at $\langle N\rangle$ and says something about it (for every $N$).