Proof of The $\delta$-Interior of a Convex Subset of $\mathbb{R}^n$ is Convex

By Convex Subset of $\mathbb{R}^n$ we must show that $t\,x+(1-t)\,y\in\Omega^{\delta}$ whenever $x,y\in\Omega^{\delta}$ and $t\in\mathbb{R}$ satisfies $0\le t$ and $t\le 1$. (If $\Omega^{\delta}$ is empty there is nothing to prove.)

Fix such $x$, $y$ and $t$ and put $z=t\,x+(1-t)\,y$. By the definition of $\Omega^{\delta}$ it suffices to prove that $\bar B(z,\delta)\subseteq\Omega$.

Let $w\in\bar B(z,\delta)$, so that $d(z,w)\le\delta$ by Closed Ball in a Metric Space, and put $v=w-z$. By claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square we have $\lVert v\rVert=\lVert w-z\rVert=d(w,z)$, and $d(w,z)=d(z,w)$ by the symmetry clause 3 of a metric; hence $\lVert v\rVert\le\delta$.

Step 1 (the translates of $x$ and $y$ lie in $\Omega$). By claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, $d(x,x+v)=\lVert v\rVert\le\delta$, so $x+v\in\bar B(x,\delta)$; since $x\in\Omega^{\delta}$ we have $\bar B(x,\delta)\subseteq\Omega$, whence $x+v\in\Omega$. The same argument gives $y+v\in\Omega$.

Step 2 ($w$ is the corresponding convex combination). In the real vector space $\mathbb{R}^{n}$ of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, distributivity of scalar multiplication over vector addition together with associativity and commutativity of addition gives

$$t\,(x+v)+(1-t)\,(y+v)=\bigl(t\,x+(1-t)\,y\bigr)+\bigl(t\,v+(1-t)\,v\bigr),$$

and distributivity of scalar multiplication over addition of scalars gives $t\,v+(1-t)\,v=\bigl(t+(1-t)\bigr)v=1\,v=v$. Hence

$$t\,(x+v)+(1-t)\,(y+v)=z+v.$$

Moreover $v=w-z=w+(-z)$ by claim 3 of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, so $z+v=z+\bigl(w+(-z)\bigr)=w+\bigl(z+(-z)\bigr)=w$ by associativity, commutativity, claim 2 of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space and claim 1 of the same result, which identifies the zero vector.

Step 3 (conclusion). By Steps 1 and 2, $w$ is the convex combination $t\,(x+v)+(1-t)\,(y+v)$ of the two points $x+v,y+v\in\Omega$; since $\Omega$ is convex, $w\in\Omega$. As $w\in\bar B(z,\delta)$ was arbitrary, $\bar B(z,\delta)\subseteq\Omega$, that is, $z\in\Omega^{\delta}$. Therefore $\Omega^{\delta}$ is convex.