The closure of an irreducible subset of an irreducible space is irreducible.

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.