Proof of The Closure is the Smallest Closed Superset

We use the definition of the closure throughout: a point $x\in X$ lies in $\operatorname{cl}_X(A)$ exactly when $U\cap A\neq\varnothing$ for every $U\in\mathcal{T}$ with $x\in U$. Note that claims 1 to 4 are proved for an arbitrary subset of $X$, so they may be applied to other subsets in the proof of claim 5.

Claim 1. Let $x\in A$ and let $U\in\mathcal{T}$ with $x\in U$. Then $x\in U\cap A$, so $U\cap A\neq\varnothing$. As $A\subseteq X$ we get $x\in\operatorname{cl}_X(A)$, and hence $A\subseteq\operatorname{cl}_X(A)$.

Claim 2. By claim 2 of Duality Between Interior and Closure Under Complementation we have $X\setminus\operatorname{cl}_X(A)=\operatorname{int}_X(X\setminus A)$, where $\operatorname{int}_X$ denotes the interior in $X$. By claim 2 of The Interior is the Largest Open Subset, applied to the subset $X\setminus A$ of $X$, the set $\operatorname{int}_X(X\setminus A)$ belongs to $\mathcal{T}$. Hence $X\setminus\operatorname{cl}_X(A)\in\mathcal{T}$, which is precisely the statement that $\operatorname{cl}_X(A)$ is closed.

Claim 3. Let $C\subseteq X$ be closed with $A\subseteq C$, and let $x\in X\setminus C$. Put $U=X\setminus C$. Then $U\in\mathcal{T}$ because $C$ is closed, and $x\in U$. Moreover $U\cap A\subseteq U\cap C=\varnothing$, so $U\cap A=\varnothing$, and therefore $x\notin\operatorname{cl}_X(A)$. Thus no point of $X\setminus C$ belongs to $\operatorname{cl}_X(A)$. Since $\operatorname{cl}_X(A)\subseteq X$, this gives $\operatorname{cl}_X(A)\subseteq C$.

Claim 4. If $A=\operatorname{cl}_X(A)$, then $A$ is closed by claim 2. Conversely, if $A$ is closed, then claim 3 applied with $C=A$ gives $\operatorname{cl}_X(A)\subseteq A$, and claim 1 gives $A\subseteq\operatorname{cl}_X(A)$; hence $A=\operatorname{cl}_X(A)$.

Claim 5. Let $B\subseteq X$ with $A\subseteq B$. By claim 1 applied to $B$ we have $B\subseteq\operatorname{cl}_X(B)$, and by claim 2 applied to $B$ the set $\operatorname{cl}_X(B)$ is closed. Hence $\operatorname{cl}_X(B)$ is a closed subset of $X$ containing $A$, and claim 3, applied to $A$ with $C=\operatorname{cl}_X(B)$, gives $\operatorname{cl}_X(A)\subseteq\operatorname{cl}_X(B)$.