Proving that the 'finite complement topology' is in fact a topology I am trying to rigorously work out the details of what Munkres calls the finite complement topology and, in particular, prove that it satisfies the axioms for a topology. There are a few details, e.g., taking a collection of nonempty sets, but the subcollection itself may be empty, that are a bit difficult to work out.
The precise statement of what I am trying to prove:

Let $X$ be a set, and $\mathcal{T}_f$ the collection of all subsets $U$ of $X$ such that $X - U$ either is finite or all of $X$. Then $\mathcal{T}_f$ is a topology on $X$, which we call the finite complement topology.

My attempt:

We have $X - X = \emptyset$, which is finite, and $X - \emptyset = X$, which is all of $X$, so $\emptyset, X \in \mathcal{T}_f$. Let $\{U_{\alpha}\}_{\alpha \in I}$ be an indexed family of nonempty elements of $\mathcal{T}_f$. It suffices to consider a collection of nonempty sets, as the empty set, while open in $\mathcal{T}_f$, contributes nothing to their union. Furthermore, if one element of this family of sets were empty, then their intersection would be the empty set, which we know to be open. We first claim that $\bigcup\limits_{\alpha \in I} U_{\alpha} \in \mathcal{T}_f$. We have
\begin{align*}
X - \bigcup\limits_{\alpha \in I} U_{\alpha} = \bigcap\limits_{\alpha \in I} \left(X - U_{\alpha}\right)
\end{align*}
by DeMorgan's law. Let $E'$ be one such $U_{\alpha}$. (If there is no such $E'$, then any $x \in X$ vacuously satisfies membership in $\bigcap\limits_{\alpha \in I} (X - U_{\alpha})$, so $\bigcap\limits_{\alpha \in I} (X - U_{\alpha}) = \emptyset$, which is open.) Then, since $E' \neq \emptyset$, $X - E' \neq X$, so $X - E'$ is finite since $E' \in \mathcal{T}_f$. But then $\bigcap\limits_{\alpha \in I} \left(X - U_{\alpha}\right) \subset X - E'$, and since any subset of a finite set is finite, we have $\bigcap\limits_{\alpha \in I} \left(X - U_{\alpha}\right)$ is finite, hence $X - \bigcup\limits_{\alpha \in I} U_{\alpha}$ is finite, so $\bigcup\limits_{\alpha \in I} U_{\alpha} \in \mathcal{T}_f$. Next, we claim that $\bigcap\limits_{\alpha \in I} U_{\alpha} \in \mathcal{T}_f$. We have
\begin{align*}
X - \bigcap\limits_{\alpha \in I} U_{\alpha} = \bigcup\limits_{\alpha \in I} (X - U_{\alpha}).
\end{align*}
For each $\alpha$, we have $U_{\alpha} \in \mathcal{T}_f$ and $U_{\alpha} \neq \emptyset$, so $X - \bigcup\limits_{\alpha \in I} (X - U_{\alpha})$ is a union of finite sets and therefore finite, so $X - \bigcap\limits_{\alpha \in I} U_{\alpha}$ is finite, so $\bigcap\limits_{\alpha \in I} U_{\alpha} \in \mathcal{T}_f$.

Am I correct that when Munkres says "any subcollection of open sets is open," this subcollection can be empty, even when we specify that the sets in the subcollection are nonempty?
 A: For unions you overcomplicate things: let $U_\alpha, \alpha \in I$ be a collection of open sets. We can assume WLOG that $I \neq \emptyset$ because otherwise the union would be empty anyway and we already know that $\emptyset \in \mathcal{T}_f$.
So $I \neq \emptyset$ and we can also assume that we have some $\alpha_0 \in I$ such that $U_{\alpha_0} \neq \emptyset$ or else all $U_\alpha$ are empty and so is the union, and we already checked $\emptyset \in \mathcal{T}_f$.
So $X-U_{\alpha_0} \neq X$ so $X-U_{\alpha_0}$ is finite. It's clear that $X-\bigcup_{\alpha \in I} U_\alpha \subseteq X-U_{\alpha_0}$ (if $x$ is not in the union, it's not in $U_{\alpha_0}$ in particular), and a subset of a finite set is finite, so $ \bigcup_{\alpha \in I} U_\alpha \in \mathcal{T}_f$.

For intersections you only have to check $U,V \in \mathcal{T}_f \to U \cap V \in \mathcal{T}_f$ and there we two trivial cases: $U=\emptyset$ and $V = \emptyset$ that we can rule out right away (then $U \cap V = \emptyset$ too) so $X-U, X-V$ are both finite and then $X-(U \cap V)= (X-U) \cup (X-V)$ is a union of two finite sets so finite and indeed $U \cap V \in \mathcal{T}_f$. QED.

In summary, once you've checked the empty set axiom, you can assume in checking the definition of a topology that in the union axiom we have a non-empty collection $U_i, i \in I$ of non-empty subsets and for the intersection axiom that we have two sets both of which are non-empty and not $X$. This can save you a litte case distinguishing in the proof.
