Let $X$ be a topological space which is a union of finitely many irreducible closed sets $X_1, ... , X_n$.
Lemma: if none of the $X_i$ are contained in one another, then $X_1, ... , X_n$ are the distinct irreducible components of $X$.
Now let $Y \subseteq X$ have finitely many irreducible components $Y_1, ... , Y_n$. Then (what I want to prove is) $\overline{Y}$ has finitely many irreducible components $\overline{Y_1}, ... , \overline{Y_n}$; and they are distinct.
Since $\overline{Y}$ is the union of irreducible closed subsets $\overline{Y_1}, ... , \overline{Y_n}$, we can apply the lemma if we can show that none of the $\overline{Y_i}$ are contained in one another. Why is this?