Definition of irreducible components in topological spaces

Let $X$ be $\mathbb{N}$ in the cofinite topology, which has as only closed sets the finite sets and $X$ itself. This is hereditary: if $A \subseteq X$, its subspace topology is also cofinite.

Any infinite cofinite space $X$ is irreducible (we can only write it as $A \cup B$ both closed and not the whole space iff $A$ and $ B$, and thus $X$ is finite.

So the even numbers in $X$ are irreducible but not maximally so (as $X$ itself is larger and also irreducible):

$A$ is an irreducible component of $X$ iff $A$ is irreducible and for every $A \subseteq B \subseteq X$: if $B$ is irreducible then $A = B$.

It's probably the case that $F \subset X$ is a maximal irreducible set if it's irreducible and the following implication holds: $$F \subset A \text{ and } A \text{ is irreducible} \implies F = A$$

Example: consider $X = \{0,1\}$ with the discrete topology. The set $\{1\}$ is an irreducible component.

Let $F$ be an irreducible component. Let's show that $\overline F$ is irreducible. Suppose not and write $\overline F = A \cup B$ where $A$ and $B$ are proper closed subsets of $\overline F$. We have $A = A' \cap \overline F$ and $B = B' \cap \overline F$ where $A'$ and $B'$ are closed in $X$. Let $A'' = A' \cap F$ and $B'' = B' \cap F$. We have that $A''$ and $B''$ are closed in $F$ and: $$A'' \cup B'' = (A' \cup B') \cap F = (A' \cup B') \cap \overline F \cap F = [(A' \cap \overline F) \cup (B' \cap \overline F)] \cap F = (A \cup B) \cap F = \overline F \cap F = F$$

Also note that if $A'' = F$, then $F \subset A'$, hence $\overline F \subset \overline A' = A'$, so $A = \overline F$, a contradiction. Similarly, we get that $A''$ and $B''$ are proper subsets of $F$. It's also evident that they're non-empty. Thus, $F$ is reducible, a contradiction. Therefore, $\overline F$ is irreducible and so, $F = \overline F$, so $F$ is closed.

To check if a subset $F \subseteq X$ is irreducible, it is considered as its own topological space via the induced topology. So we isolate the set $F$, give it the induced topology (forgetting a lot of stuff about $X$!), and check if that new space is irreducible. But it might be the case that while $F$ considered as a space is irreducible, $F \cup G$ is also irreducible for some other subset $G \subseteq X$.

As an example, take the set $X = \{0, \ldots, n\}$ with the coarse topology where the only closed sets are $\emptyset$ and $X$. Here any subset $F \subseteq X$ will be coarse again, and hence irreducible, and so a maximal such subset is $X$. You can make a similar example using the Zariski topology: circle part of an irreducible curve on a page: that circled chunk should be irreducible as a topological space, but it is certainly not maximal.