[proofplan] We prove the contrapositive obstruction: if a nonempty open subset $U$ failed to be dense, then its closure would be a proper closed subset of $X$. The complement $X\setminus U$ is also a proper closed subset because $U$ is open and nonempty. Since $U\subset \overline{U}$, these two proper closed subsets cover $X$, contradicting irreducibility. [/proofplan] [step:Assume a nonempty open subset has proper closure] Let $U\subset X$ be nonempty and open. Suppose, for contradiction, that $U$ is not dense in $X$. By the definition of density, this means that its closure $\overline{U}$ is not all of $X$, so $\overline{U}\subsetneq X$. By the definition of closure, $\overline{U}$ is closed in $X$ and contains $U$. [/step] [step:Cover the space by two proper closed subsets] Since $U$ is open in $X$, the complement $X\setminus U$ is closed in $X$. Since $U$ is nonempty, $X\setminus U$ is a proper subset of $X$. By the previous step, $\overline{U}$ is also a proper closed subset of $X$. We claim that $X=(X\setminus U)\cup \overline{U}$. Indeed, let $x\in X$. If $x\notin U$, then $x\in X\setminus U$. If $x\in U$, then $x\in \overline{U}$ because $U\subset \overline{U}$. Thus every point of $X$ lies in $(X\setminus U)\cup \overline{U}$, and the reverse inclusion is immediate because both sets are subsets of $X$. [guided] The contradiction comes from turning the failure of density into a forbidden decomposition of $X$. Since $U$ is open, its complement $X\setminus U$ is closed. This complement is proper because $U$ is nonempty: choose any point $u\in U$, and then $u\notin X\setminus U$. The other [closed set](/page/Closed%20Set) is $\overline{U}$. By definition, $\overline{U}$ is closed and contains $U$. Our contradiction hypothesis says $\overline{U}\ne X$, so $\overline{U}$ is also a proper closed subset of $X$. Now we verify the covering identity point by point: $X=(X\setminus U)\cup \overline{U}$. Let $x\in X$. There are two cases. If $x\notin U$, then $x\in X\setminus U$. If $x\in U$, then $x\in \overline{U}$ because the closure of a set contains the set itself. Hence every $x\in X$ belongs to $(X\setminus U)\cup \overline{U}$. Since both $X\setminus U$ and $\overline{U}$ are subsets of $X$, the reverse inclusion also holds. [/guided] [/step] [step:Use irreducibility to force density] The preceding step writes $X$ as the union of two proper closed subsets: \begin{align*} X=(X\setminus U)\cup \overline{U}. \end{align*} This contradicts the irreducibility of $X$, which says that $X$ cannot be expressed as the union of two proper closed subsets. Therefore the assumption $\overline{U}\subsetneq X$ is impossible. Hence $\overline{U}=X$, so $U$ is dense in $X$. [/step]