Why is the class of recursively enumerable languages not closed under complementation?
The statement you quote is an argument why the proof for showing that $\mathrm{R}$ (the set of recursive languages) is closed against complement does not work for $\mathrm{RE}$ (the set of recursively enumerable languages). It's not a proof of the reverse in itself -- another proof may still work!

The problem is that flipping the result of a TM only works if you get a result; $\mathrm{RE}$ promises only an acceptor, not a decider, which may loop if the input is not in the language. Therefore, you can not flip properly; all you'd get is a machine that says "no" of the word is not the language but loops otherwise. That's a machine for $\mathrm{co\text{-}RE}$, not $\mathrm{RE}$.

As for why $\mathrm{RE}$ is not closed against complement, the site you link contains all the answers; let me reiterate.


If $L$ and $\overline{L}$ are both semi-decidable, then both are decidable.
Hence, if $\mathrm{RE}$ was closed against complement, we'd have $\mathrm{R} = \mathrm{RE}$.
But we know that $\mathrm{R} \subsetneq \mathrm{RE}$ (by witness of the halting problem), so this can't be true.