If $ Y $ is irreducible set so is $cl(Y)$. If $ Y $ is irreducible set so is $cl(Y)$.
If $cl(Y)$ is reducible then $cl(Y)= A \cup B$ where both $A$ and $B$ is closed in $cl(Y)$. Now how do we proceed?

Suppose $Y$ is an irreducible subset of a larger topological space $X$.  If $\overline{Y} = A \cup B$, with $A$ and $B$ closed sets of $\overline{Y}$, then $A$ and $B$ are actually closed sets in $X$.  Then $$Y = Y \cap \overline{Y} = Y \cap (A \cup B) = (Y \cap A) \cup (Y \cap B)$$ with $Y \cap A$ and $Y \cap B$ closed subsets of $Y$.  Since $Y$ is irreducible, one of these closed sets, say $Y \cap A$, is equal to $Y$.  Thus $Y \subseteq A$, and hence $\overline{Y} \subseteq A$.  But then $\overline{Y} = A$.  
