[proofplan]
We verify the topology axioms directly. The empty set belongs to $\tau_{\mathrm{cof}}$ by definition, and $X$ belongs because its complement in $X$ is $\varnothing$. For arbitrary unions, either every member is empty or one member has finite complement, which forces the union to have finite complement. For finite intersections, we separate off the case where one factor is empty, and otherwise use De Morgan's law and the fact that a finite union of finite sets is finite.
[/proofplan]
[step:Verify that $\varnothing$ and $X$ belong to $\tau_{\mathrm{cof}}$]
By definition of $\tau_{\mathrm{cof}}$, we have $\varnothing \in \tau_{\mathrm{cof}}$.
Also,
X \setminus X = \varnothing.
Since $\varnothing$ is finite, the defining condition for membership in $\tau_{\mathrm{cof}}$ gives $X \in \tau_{\mathrm{cof}}$.
[/step]
[step:Show that arbitrary unions of members of $\tau_{\mathrm{cof}}$ remain in $\tau_{\mathrm{cof}}$]
Let $I$ be an index set, and let $(U_i)_{i \in I}$ be a family of subsets of $X$ such that $U_i \in \tau_{\mathrm{cof}}$ for every $i \in I$. Define
U := \bigcup_{i \in I} U_i.
If $U_i = \varnothing$ for every $i \in I$, then $U = \varnothing$, so $U \in \tau_{\mathrm{cof}}$.
Otherwise, there exists $j \in I$ such that $U_j \neq \varnothing$. Since $U_j \in \tau_{\mathrm{cof}}$ and $U_j \neq \varnothing$, the set $X \setminus U_j$ is finite. Because $U_j \subset U$, taking complements in $X$ gives
\begin{align*}
X \setminus U \subset X \setminus U_j.
\end{align*}
A subset of a finite set is finite, so $X \setminus U$ is finite. Therefore $U \in \tau_{\mathrm{cof}}$.
[guided]
Let $I$ be an index set, and let $(U_i)_{i \in I}$ be a family of subsets of $X$ with $U_i \in \tau_{\mathrm{cof}}$ for each $i \in I$. We must prove that the union
U := \bigcup_{i \in I} U_i.
also belongs to $\tau_{\mathrm{cof}}$.
There are two cases because $\varnothing$ was included separately in the definition of $\tau_{\mathrm{cof}}$. If every $U_i$ is empty, then
U = \bigcup_{i \in I} \varnothing = \varnothing,
and hence $U \in \tau_{\mathrm{cof}}$ by definition.
Now suppose that at least one member of the family is nonempty. Choose an index $j \in I$ such that $U_j \neq \varnothing$. Since $U_j \in \tau_{\mathrm{cof}}$ and $U_j$ is not the separately included empty set case, the defining condition gives that $X \setminus U_j$ is finite.
The key point is that an arbitrary union is larger than each of its members. In particular,
U_j \subset U.
Taking complements inside the fixed ambient set $X$ reverses inclusion, so
\begin{align*}
X \setminus U \subset X \setminus U_j.
\end{align*}
Since $X \setminus U_j$ is finite and every subset of a finite set is finite, $X \setminus U$ is finite. Therefore $U$ satisfies the cofinite membership condition, so $U \in \tau_{\mathrm{cof}}$.
[/guided]
[/step]
[step:Show that finite intersections of members of $\tau_{\mathrm{cof}}$ remain in $\tau_{\mathrm{cof}}$]
Let $n \in \mathbb{N}$, and let $U_1,\dots,U_n \in \tau_{\mathrm{cof}}$. Define
V := \bigcap_{k=1}^{n} U_k.
If $U_k = \varnothing$ for some $k \in \{1,\dots,n\}$, then $V = \varnothing$, so $V \in \tau_{\mathrm{cof}}$.
Otherwise, $U_k \neq \varnothing$ for every $k \in \{1,\dots,n\}$. Since each $U_k \in \tau_{\mathrm{cof}}$, each complement $X \setminus U_k$ is finite. By De Morgan's law,
X \setminus V = X \setminus \bigcap_{k=1}^{n} U_k = \bigcup_{k=1}^{n} (X \setminus U_k).
The right-hand side is a finite union of finite sets, hence finite. Therefore $X \setminus V$ is finite, so $V \in \tau_{\mathrm{cof}}$.
[/step]
[step:Conclude that $\tau_{\mathrm{cof}}$ satisfies the topology axioms]
We have shown that $\varnothing,X \in \tau_{\mathrm{cof}}$, that $\tau_{\mathrm{cof}}$ is closed under arbitrary unions, and that $\tau_{\mathrm{cof}}$ is closed under finite intersections. These are exactly the axioms for a topology on $X$. Hence $\tau_{\mathrm{cof}}$ is a topology on $X$.
[/step]
