Convex Set with Empty Interior Lies in an Affine Set

Look at $d+1$, the largest number of affinely independent points from $C$. Let $x_0$, $\ldots$, $x_d$ one such affinely independent subset of largest size. Note that every other point is an affine combination of the points $x_k$, so lies in the affine subspace generated by them, which is of dimension $d$.

If $d < n$ then this subspace is contained in an affine hyperplane.

If $d=n$, then $C$ contains $d+1$ affinely independent points. Since $C$ is convex, it will also contain the convex hull of those $n+1$ points. Now, in an $n$-dimensional space the convex hull of $n+1$ affinely independent points has non-empty interior. So the interior of $C$ is non-empty.

Let $C$ be a convex with empty interior of $R^n$. Suppose that $C$ contains two points $x_1,x_2$ then it contains the segment $[x_1,x_2]$, if $C$ is not contained in the affine line $D_2$ which contains $x_1,x_2$ then there exists an element $x_3$ of $C$ not in $D_2$, thus the 2 simplex $[x_1,x_2,x_3]$ is contained in $C$, recursively suppose constructed a $i-1$-simplex $[x_1,...,x_i]\subset C$, if $C$ is not contained in the affine $i-1$-plane $D_i$ which contains $[x_1,...,x_i]$, you have $x_{i+1}\in C$ not in the $i-1$-plan $D_i$, thus $C$ contains $[x_1,....,x_{i+1}]$. This shows that if $C$ is not contained in an affine subspace of $R^n$ distinct of $R^n$, then it contains a $n$ simplex $[x_1,...,x_{n+1}]$ so its interior is not empty.