Show "countable complement topology" is a topology

The proofs are almost OK. You do need to handle the special case of the empty set in your proofs.

So for (ii): suppose $U_1,\ldots,U_n$ are in $\mathscr{T}$, where $n \in \mathbb{N}$.

If one of the $U_i$ happens to be equal to $\emptyset$, then $\cap_{i=1}^n U_i = \emptyset$ as well so $\cap_{i=1}^n U_i \in \mathscr{T}$ as required.

In the other case, all $U_i$ obey that $X \setminus U_i$ is at most countable, so then de Morgan says that $X \setminus (\cap_{i=1}^n U_i) = \cup_{i=1}^n (X \setminus U_i)$, which is a finite union of (at most) countable sets so (at most) countable. Hence $\cap_{i=1}^n U_i \in \mathscr{T}$, as required.

As an aside: Note that in this case, because we know that even a countable union of (at most) countable sets is (at most) countable, we can even show that the countable intersection of open sets is open. Also the intersection of (countably many) non-empty open sets will be non-empty, in case $X$ is uncountable.

As to (iii): if $U_i, i \in I$ are in $\mathscr{T}$, then we can assume all $U_i \neq \emptyset$, because they do not contribute to the union, and we can assume we have at least one $U_{i_0}$ that is non-empty (or the union would be empty and so in $\mathscr{T}$ anyway).

Now, $X \setminus ( \cup_{i \in I} U_i ) = \cap_{i \in I} (X \setminus U_i) \subseteq X \setminus U_{i_0}$, and the last set is at most countable, so $\cup_{i \in I} U_i \in \mathscr{T}$, being co-countable as well.

I think that (as this is a topology course) claiming that a finite (or countable) union of countable sets is countable, is quite justified without giving a proof for that: such basic set theory should be a given before a topology course, in my opinion.