Proof of Euclidean Balls are Convex

Write $\lVert\,\cdot\,\rVert$ for the Euclidean norm and $|\,\cdot\,|$ for the absolute value on $\mathbb{R}$. Claims 2, 5 and 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ are used as the identity $d_E(u,v)=\lVert u-v\rVert$, homogeneity $\lVert\mu u\rVert=|\mu|\,\lVert u\rVert$, and the triangle inequality for the norm. Order arithmetic is taken from Elementary Arithmetic in an Ordered Field (claim 3 for translation of an inequality, claim 5 for multiplication by a nonnegative element) and from Elementary Order Arithmetic in an Ordered Field (claim 1 for strict compatibility with addition, claim 2 for mixed transitivity, claim 10 for strict compatibility with multiplication by a positive element). Points of $\mathbb{R}^n$ are added, scaled and subtracted coordinatewise by Sum of Points of $\mathbb{R}^n$, Scalar Multiple of a Point of $\mathbb{R}^n$ and Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$, and $\mathbb{R}^n$ is a real vector space by Euclidean Space $\mathbb{R}^n$ is a Real Vector Space.

Let $x,y$ lie in the set under consideration, let $t\in\mathbb{R}$ satisfy $0\le t$ and $t\le1$, and put $z=t\,x+(1-t)\,y$. Translating $t\le 1$ by $-t$ gives $0\le 1-t$.

Step 1. $z-a=t\,(x-a)+(1-t)\,(y-a)$. Indeed, for each index $i$ distributivity gives

and $t\,a_i+(1-t)\,a_i=\bigl(t+(1-t)\bigr)a_i=1\,a_i=a_i$ by distributivity and the field axioms of the field $\mathbb{R}$, so the two points agree in every coordinate.

Step 2. Using the identity for $d_E$, the triangle inequality, homogeneity, and $|t|=t$ and $|1-t|=1-t$, which hold because $0\le t$ and $0\le 1-t$,

where the last equality also uses the symmetry axiom of Metric Space. Note further that $t\,r+(1-t)\,r=\bigl(t+(1-t)\bigr)r=r$.

Claim 2 (closed ball). Suppose $d_E(a,x)\le r$ and $d_E(a,y)\le r$. Multiplying by the nonnegative elements $t$ and $1-t$ gives $t\,d_E(a,x)\le t\,r$ and $(1-t)\,d_E(a,y)\le(1-t)\,r$. Adding the first inequality translated by $(1-t)\,d_E(a,y)$ to the second translated by $t\,r$, and using transitivity of $\le$, gives

so $d_E(a,z)\le r$ by step 2 and transitivity. Hence $z\in\overline{B}(a,r)$.

Claim 1 (open ball). Suppose $d_E(a,x)<r$ and $d_E(a,y)<r$. If $t=0$ then $z=0\,x+1\,y=y$, and if $t=1$ then $z=1\,x+0\,y=x$, by the vector space axioms; in both cases $z$ lies in $B_{d_E}(a,r)$. Otherwise $0<t$, and $t\ne1$ together with $t\le1$ gives $t<1$, whence $0<1-t$ by translation by $-t$. Strict compatibility with multiplication by a positive element gives

Adding $(1-t)\,d_E(a,y)$ to the first and $t\,r$ to the second, then using mixed transitivity, gives

and step 2 with mixed transitivity gives $d_E(a,z)<r$. Hence $z\in B_{d_E}(a,r)$.

In both cases the defining condition of Convex Subset of $\mathbb{R}^n$ holds, so both sets are convex.