Chapter 1, Exercise 1.6: Subsets and closures of irreducible spaces

$\renewcommand{\bar}{\overline}$ Any nonempty open subset of an irreducible topological space is dense and irreducible. If $Y$ is a subset of a topological space $X$, which is irreducible in its induced topology, then the closure $\bar Y$ is also irreducible.

This is just a topology problem, and even then is not a particularly difficult one. For the first statement, we will show separately that an open $Y\subseteq X$ is dense and that it is irreducible. For density, note that $\bar Y$ and $Y^c$ are both closed and $\bar Y\cup Y^c=X$, so by irreducibility $X$ equals one of the factors: $\bar Y=X$ or $Y^c=X$. The latter implies $Y=\emptyset$ which we assume is false, thus $\bar Y=X$ and $Y$ is dense. Now, suppose we could write $Y=Y_1\cup Y_2$ where each $Y_i$ is closed. Then $X=Y_1\cup Y_2\cup Y^c$, all of which are closed, so again since $Y$ is nonempty we have $X=Y_1$ or $X=Y_2$. In either case $Y=X$ which is irreducible, so $Y=Y_1$ or $Y=Y_2$.

Now, let $Y\subset X$ be irreducible in the subspace topology and let $\bar Y = Y_1\cup Y_2$, both of which are closed. A simple calculation shows $$(Y_1\cap Y)\cup(Y_2\cap Y)=Y,$$ and both are closed in the subspace topology on $Y$. Thus $Y=Y\cap Y_1$ or $Y=Y\cap Y_2$, meaning that, without loss of generality, $Y\subset Y_1$. Then $$Y_1=Y_1\cap\bar Y=\bar{Y_1\cap Y}=\bar Y$$ as desired.