Proof of Duality Between Interior and Closure Under Complementation

Claim 1. Let $x\in X$. By the definition of the interior, the statement $x\notin\operatorname{int}_X(A)$ says that there is no $U\in\mathcal{T}$ with $x\in U$ and $U\subseteq A$; equivalently, every $U\in\mathcal{T}$ with $x\in U$ satisfies $U\not\subseteq A$. For $U\subseteq X$, the failure $U\not\subseteq A$ means that there is a point $y\in U$ with $y\notin A$; since $y\in U\subseteq X$, such a $y$ lies in $X\setminus A$, and conversely any $y\in U\cap(X\setminus A)$ witnesses $U\not\subseteq A$. So $U\not\subseteq A$ is equivalent to $U\cap(X\setminus A)\neq\varnothing$.

Consequently, for $x\in X$ the statement $x\in X\setminus\operatorname{int}_X(A)$ is equivalent to: $U\cap(X\setminus A)\neq\varnothing$ for every $U\in\mathcal{T}$ with $x\in U$. By the definition of the closure applied to the subset $X\setminus A$ of $X$, this last statement is exactly $x\in\operatorname{cl}_X(X\setminus A)$. Since both $X\setminus\operatorname{int}_X(A)$ and $\operatorname{cl}_X(X\setminus A)$ are subsets of $X$, this proves the asserted equality.

Claim 2. Because $A\subseteq X$ we have $X\setminus(X\setminus A)=A$. Applying claim 1 with $X\setminus A$ in place of $A$ therefore gives

Both $\operatorname{int}_X(X\setminus A)$ and $\operatorname{cl}_X(A)$ are subsets of $X$, so taking complements relative to $X$ on both sides of this equality yields

which is the assertion.