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 $cl(Y)$ is reducible so that $cl(Y)=A\cup B$. Then $Y\subset A\cup B$ so $Y\subset A$ or $Y\subset B$. Assume the former. But $A$ is closed so $cl(Y)\subset A$ and $A\subset cl(Y)$ so $A=cl(Y)$.
