From Hartshorne:
If $Y$ is an irreducible subset of $X$, then its closure $\overline{Y}$ in $X$ is also irreducible.
By irreducible they mean that $Y$ cannot be written $Y_1 \cup Y_2$ for two proper subsets $Y_i$ of $Y$ that are closed in the subspace topology of $Y$.
My attempt
Suppose that $\overline{Y} = Y_1 \cup Y_2$, both $Y_1, Y_2$ closed in $\overline
{Y}$. Since the closed subsets of $\overline{Y}$ are precisely the closed subsets of $X$ intersected with $\overline{Y}$, we have that $Y_i = \overline{Y} \cap Y_i'$ for some closed $Y_i'$ in $X$. In other words, each $Y_i$ is closed in the whole space $X$ as well as in $\overline{Y}$.
By a previous statement, If $U \subset Y$ is an open subset, then when $Y$ is irreducible, $U$ is both irreducible (in $U$) and dense (in $Y$).
So we have that $U_i = \overline{Y} \setminus Y_i$ is irreducible in itself and dense in $\overline{Y}$, meaning $\overline{U}_i = \overline{Y} \setminus \text{Int} (Y_i) = \overline{Y}$.
Where to next?
Best Answer
If $\overline{Y} = Y_1 \cup Y_2$, then $Y = (Y_1 \cap Y) \cup (Y_2 \cap Y)$