The closure of an irreducible subset of an irreducible space is irreducible. Start with an irreducible space $X$. Take a subset $Y$ that is irreducible. Show that the closure of $Y$ is still irreducible. 

I imagine we are supposed to start with saying, assume we have a decomposition for $\bar Y = S\cup T$ and then somehow derive a contradiction to $X$ or $Y$'s irreducibility, but I am struggling.
Suppose $\bar Y$ is not irreducible, so we have closed subsets $S,T$ in $X$ such that $\bar Y$ is not contained in either $S$ or $T$ but $\bar Y \subseteq S\cup T$. If $Y\subseteq S$, then $\bar Y\subseteq \bar S=S$, a contradiction, so $Y$ is not contained in $S$. Similarly $Y$ is not contained in $T$. But $Y\subseteq \bar Y\subseteq S\cup T$, thus $Y$ is not irreducible.
Consider $S^c$ and $T^c$, which are open in $\bar{Y}$. Then, since any nonempty open subset of an irreducible space ($Y$) is dense, we have, $\bar{S^c\bigcap Y}=Y$, and $\bar{T^c\bigcap Y}=Y$. Then, $Y=\bar{(S^c\bigcap Y})\bigcup(\bar{T^c\bigcap Y})$, which is a contradiction to Y being irreducible.