Isn't this proof of a theorem about the closedness of a set wrong?

No problems here. Note that the author doesn't assume $A^c$ to be closed. Just that it isn't open. And since "open" means "for any point in the set, we have a certain property", we get that "not open" means "for at least one point in the set, we don't have that property". The point $x$ they choose is one such point guaranteed to exist by that assumption.

There are four possiblities

1) $A^c$ is open and closed.

2) $A^c$ is open and not closed.

3) $A^c$ is not open and closed.

4) $A^c$ is not open and not closed.

He assumed: $A^c$ is not open. That means either $3$ or $4$. He got a contradiction. That shows that both $3$ and $4$ are both false. Thus the only possibilities are $1$ or $2$. So he concludes either $1$ or $2$ is true. Which means.... $A^c$ is open.

However as he didn't give a carnivore's fig whether $A^c$ was closed or not he made no mention of it at all.

So it comes down to:

I) $A^c$ is open. or

II) $A^c$ is not open.

He assumes II) gets a contradiction, and concludes I). $A^c$ is open. Now maybe $A^c$ is closed. Or maybe it isn't. He doesn't care and made no assumptions to the fact.

But he does know, and correctly so. That $A^c$ is open....