Proof of The Interior is the Largest Open Subset

We use the definition of the interior throughout: a point $x\in X$ lies in $\operatorname{int}_X(A)$ exactly when there is $U\in\mathcal{T}$ with $x\in U$ and $U\subseteq A$.

Claim 1. Let $x\in\operatorname{int}_X(A)$ and pick $U\in\mathcal{T}$ with $x\in U$ and $U\subseteq A$. Then $x\in A$. Hence $\operatorname{int}_X(A)\subseteq A$.

Claim 3. Let $U\in\mathcal{T}$ with $U\subseteq A$, and let $y\in U$. Then $U$ itself satisfies $U\in\mathcal{T}$, $y\in U$ and $U\subseteq A$, so $y\in\operatorname{int}_X(A)$. Hence $U\subseteq\operatorname{int}_X(A)$.

Claim 2. Let $\mathcal{F}$ be the collection of all $U\in\mathcal{T}$ with $U\subseteq A$, and regard $\mathcal{F}$ as an index set for the family of subsets of $X$ that assigns to each $U\in\mathcal{F}$ the set $U$ itself. We show

If $x$ lies in the union, then $x\in U$ for some $U\in\mathcal{F}$, and $U\subseteq\operatorname{int}_X(A)$ by claim 3, so $x\in\operatorname{int}_X(A)$. Conversely, if $x\in\operatorname{int}_X(A)$, pick $U\in\mathcal{T}$ with $x\in U$ and $U\subseteq A$; then $U\in\mathcal{F}$ and $x\in U$, so $x$ lies in the union.

Every member of the family belongs to $\mathcal{T}$ by construction, so by condition 2 in the definition of a topological space the displayed union belongs to $\mathcal{T}$. Hence $\operatorname{int}_X(A)\in\mathcal{T}$.

Claim 4. If $A=\operatorname{int}_X(A)$, then $A\in\mathcal{T}$ by claim 2. Conversely, if $A\in\mathcal{T}$, then applying claim 3 to $U=A$, which satisfies $A\subseteq A$, gives $A\subseteq\operatorname{int}_X(A)$; combined with claim 1 this gives $A=\operatorname{int}_X(A)$.

Claim 5. Let $B\subseteq X$ with $A\subseteq B$, and let $x\in\operatorname{int}_X(A)$. Pick $U\in\mathcal{T}$ with $x\in U$ and $U\subseteq A$. Then $U\subseteq B$, so $x\in\operatorname{int}_X(B)$. Hence $\operatorname{int}_X(A)\subseteq\operatorname{int}_X(B)$.