Proof of A Local Minimum of a Convex Function on a Convex Subset of $\mathbb{R}^n$ Is a Global Minimum

Let $y\in C$. If $y=x$ there is nothing to prove, so let $y\neq x$; then $\lVert y-x\rVert>0$ (a background fact of the setting).

Put

$$t=\min\Bigl\{1,\ \frac{r}{2\lVert y-x\rVert}\Bigr\},\qquad z=(1-t)x+ty,$$

so that $0<t\le1$ and $t\lVert y-x\rVert\le r/2<r$. The point $z$ lies in $C$ by Segments and Convex Subsets of $\mathbb{R}^n$ (ii), and $z-x=t(y-x)$, so by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (ii) and Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v)

$$d_{E}(x,z)=\lVert z-x\rVert=t\lVert y-x\rVert<r .$$

Hence $z\in C\cap B(x,r)$ by Open Ball in a Metric Space (ii).

The hypothesis at $z$ and the convexity inequality of Convex Function on a Convex Subset of $\mathbb{R}^n$ (i) now give

$$f(x)\le f(z)\le(1-t)f(x)+tf(y),$$

so $t f(x)\le t f(y)$, and $f(x)\le f(y)$ because $t>0$.