Proof of Open Ball in a Metric Space is Open

$\mathbb{R}$ is an ordered field by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals, hence an ordered ring by Ordered Fields §ordered-field and a commutative ring by Fields §field. By Open Ball in a Metric Space §ball, $B_{d}(x,r)$ is a subset of $X$. Let $y\in B_{d}(x,r)$; by Open Ball in a Metric Space §membership, $y\in X$ and $d(x,y)<r$. Put $\varepsilon=r-d(x,y)$. By Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares §difference, $0<\varepsilon$. By Negatives, Differences, Reciprocals and Quotients §negative and the commutative, associative, inverse and neutral laws of Commutative Rings §ring,

$$d(x,y)+\varepsilon=d(x,y)+\bigl((-d(x,y))+r\bigr)=\bigl(d(x,y)+(-d(x,y))\bigr)+r=0+r=r .$$

Since $\varepsilon>0$, the open ball $B_{d}(y,\varepsilon)$ is defined by Open Ball in a Metric Space §ball. Let $z\in B_{d}(y,\varepsilon)$; by Open Ball in a Metric Space §membership, $z\in X$ and $d(y,z)<\varepsilon$. Applying Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum to $d(y,z)<\varepsilon$ and $d(x,y)\le d(x,y)$, and using the commutativity of $+$ in Commutative Rings §ring, gives $d(x,y)+d(y,z)<d(x,y)+\varepsilon=r$. By Metric Space §triangle, $d(x,z)\le d(x,y)+d(y,z)$, so Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed gives $d(x,z)<r$, and $z\in B_{d}(x,r)$ by Open Ball in a Metric Space §membership. Hence $B_{d}(y,\varepsilon)\subseteq B_{d}(x,r)$ by Subclasses and Subsets §subclass.

Thus every point $y$ of $B_{d}(x,r)$ has a real number $\varepsilon>0$ with $B_{d}(y,\varepsilon)\subseteq B_{d}(x,r)$, and $B_{d}(x,r)$ is open in $(X,d)$ by Open Subset of a Metric Space §open.