Is $\mathbb{R}$ with the cofinite topology compact? To begin with I already know any set with the cofinite topology is compact, but I am looking for a proof specific to $\mathbb{R}_{\mathcal{FC}}$. My attempt to a proof is right here:
Recall that any $A \subset \mathbb{R}_{\mathcal{FC}}$ is open if $A=\varnothing$ or $\mathbb{R}\setminus A$ is finite and as a result every closed $X\subset\mathbb{R}_{\mathcal{FC}}$ is finite or $X=\mathbb{R}$ making $\mathbb{R}$ closed. By definition $\mathbb{R}$ is also open, so $\mathbb{R}$ is clopen. Recall that also $\varnothing$ is open by definition but also closed as it is finite, so it too is the only other clopen set.
If every closed set is finite in $\mathbb{R}_{\mathcal{FC}}$ then every open set must be infinite. As a result every possible open cover must the union of infinite sets in which case a subcover could be expressed as a finite union of infinite sets, so $\mathbb{R}_{\mathcal{FC}}$ is compact.
 A: I don't trust this sentence.

As a result every possible open cover must the union of infinite sets in which case a subcover could be expressed as a finite union of infinite sets, so $\Bbb{R}_\mathcal{FC}$ is compact.

More convincing would be to pick any nonempty member of the cover, $U$, label the points in its finite complement, $x_1, \dots, x_n$, then observe that each such point is covered by at least one member of the cover, $U(x_i)$.  Then take as a finite cover $\{U, U(x_1), \dots, U(x_n)\}$.
A: To see why your argument fails, consider the cocountable topology on $\mathbb R$ where a non-empty set is open iff its complement is countable.
Since every non-empty open set is the complement of a countable set, it is uncountable, and therefore infinite. As a result, again every possible open cover must be the union of infinite sets (and possibly the empty set, but that's true also for the cofinite metric). Therefore if your argument were true, $\mathbb R$ under the cocountable topology would have to be compact as well.
Now consider the open cover $\mathcal C=\{U_n\mid n\in\mathbb N\}$ with $U_n=(\mathbb R\setminus\mathbb N)\cup\{n\}$. For every finite subcover $\mathcal F\subset\mathcal C$, there is some $k\in\mathbb N$ such that $U_k\notin\mathcal F$. But then $k$ is not covered by $\mathcal F$. Therefore $\mathbb R$ under the cocountable topology is not compact, in contradiction to what your argument would say if it were valid.
Therefore your argument is not valid.
A: To see why it is compact, let $\{U_a\}_{a \in A}$, $A$ some arbitrary indexing set, be an open cover for $\Bbb{R}$. Then for some $b \in A$, we have for each $x \in \Bbb{R} \setminus U_b$ an open set $U_x$ containing $x$, then
$$\Bbb{R} \subset U_b \cup \bigcup_{x \in \Bbb{R} \setminus U_b}U_x$$
which is finite as $\Bbb{R} \setminus U_b$ is finite.
