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

Let $(X,d)$ be a metric space and let $\mathcal{T}_{d}$ be the collection of the subsets of $X$ that are open in the metric space $(X,d)$; by Metric Open Sets Form a Topology the pair $(X,\mathcal{T}_{d})$ is a topological space, and interiors below are taken in it, in the sense of Interior of a Subset of a Topological Space. Open balls $B_{d}(x,r)$ are those of Open Ball in a Metric Space and closed balls $\bar{B}_{d}(x,s)$ those of Closed Ball in a Metric Space. Let $\mathbb{R}$ be the set of real numbers with the order $\le$ of its ordered field structure, and for $a,b\in\mathbb{R}$ write $a<b$ to mean that $a\le b$ and $a\ne b$.

Let $A\subseteq X$ and let $x\in X$. Then the following hold.

1. (An interior point is the centre of a contained closed ball) If $x\in\operatorname{int}_{X}(A)$, then there is $s\in\mathbb{R}$ with $0<s$ and

$$\bar{B}_{d}(x,s)\subseteq A .$$

2. (Conversely) If there is $s\in\mathbb{R}$ with $0<s$ and $\bar{B}_{d}(x,s)\subseteq A$, then $x\in\operatorname{int}_{X}(A)$.