Proof of Interior Points in the Metric Topology are Exactly the Centres of Contained Closed Balls

Claim 1. Let $x\in\operatorname{int}_{X}(A)$. By Interior of a Subset of a Topological Space there is $U\in\mathcal{T}_{d}$ with $x\in U$ and $U\subseteq A$. Since $U$ is open in the metric space $(X,d)$, Open Subset of a Metric Space provides $r\in\mathbb{R}$ with $0<r$ and $B_{d}(x,r)\subseteq U$.

Put $s=r\,2^{-1}$, where $2=1+1$. By claim 8 of Elementary Order Arithmetic in an Ordered Field we have $0<2$, so $2^{-1}$ exists and $0<2^{-1}$ by claim 7 of that lemma, and $0<s$ by claim 5. Moreover $s+s=r(2^{-1}+2^{-1})=r$, so adding $s$ to the inequality $0<s$ by claim 1 of Elementary Order Arithmetic in an Ordered Field gives $s<r$.

Let $y\in\bar{B}_{d}(x,s)$, so that $d(x,y)\le s$ by Closed Ball in a Metric Space. Together with $s<r$ and the mixed transitivity of claim 2 of Elementary Order Arithmetic in an Ordered Field this gives $d(x,y)<r$, that is, $y\in B_{d}(x,r)$ by Open Ball in a Metric Space. Hence

Claim 2. Let $s\in\mathbb{R}$ with $0<s$ and $\bar{B}_{d}(x,s)\subseteq A$. If $y\in B_{d}(x,s)$ then $d(x,y)<s$, hence $d(x,y)\le s$ and $y\in\bar{B}_{d}(x,s)$; so

By Open Ball in a Metric Space is Open the set $B_{d}(x,s)$ is open in the metric space $(X,d)$, so it belongs to $\mathcal{T}_{d}$. By Metric Space we have $d(x,x)=0$, and $0<s$, so $x\in B_{d}(x,s)$.

Thus $B_{d}(x,s)$ is a member of $\mathcal{T}_{d}$ containing $x$ and contained in $A$, and Interior of a Subset of a Topological Space gives $x\in\operatorname{int}_{X}(A)$.