I'd like to proof the following statement:
Let $A, B \subset \mathbb{R}^n$ both be convex and closed. Additionally, let $A$ be full-dimensional (e.g. $\text{int}(A) \neq \emptyset$). Then $\text{int}(A+B) = \text{int}(A) + B$.
The inclusion $\text{int}(A) + B \subset \text{int}(A+B)$ is rather easy to proof, but I cannot figure out the other direction, even though it seems like a rather intuitive statement.
Proposition. Let $A$ be a convex set with $a_0\in {\rm int}(A)$ and $a_1\in A$. Then $a_\theta := a_1 + \theta(a_0-a_1) \in {\rm int}(A)$ for all $\theta\in (0,1)$.
Proof. $a_0\in {\rm int}(A)$ means there is a ball $B(a_0, \epsilon) \in A$ with $\epsilon > 0$. Convexity implies that $B(a_\theta, \theta\epsilon) \in A$, this $a_\theta \in {\rm int}(A)$. $\Box$
Now we aim to show ${\rm int}(A+B) \subseteq {\rm int}(A) + B$. Let $x\in {\rm int}(A+B)$. This implies two things:
Since ${\rm int}(A) \neq \emptyset$, let $a_0$ be any point in ${\rm int}(A)$. Define $v := \frac{\epsilon(a-a_0)}{2|a-a_0|}$. $x+v = (a+v)+b$, so $x+v\in A+B$. By the proposition above, $a+v \in {\rm int}(A)$.
Also, $x-v\in B(x,\epsilon)\subseteq A+B$ so there exist $a_{-}\in A$ and $b_{-}\in B$ such that $x-v = a_{-} + b_{-}$. Then, \begin{align} x &= \frac{(x+v) + (x-v)}{2}\\ &= \frac{(a+v + b) + (a_{-}+b_{-})}{2}\\ &= \frac{a+v + a_{-}}{2} + \frac{b+b_{-}}{2}. \end{align}
Convexity of $B$ implies that $\frac{b+b_{-}}{2}\in B$. Since $a+v \in {\rm int}(A)$, the proposition above implies that $\frac{a+v + a_{-}}{2} \in {\rm int}(A)$. Thus $x\in {\rm int}(A) + B$.