Radstrom cancellation only for two convex sets? I've seen this statement of Radstrom cancellation:
if $A +C \subset B+C$ where $A,B$ are convex, $B$ is closed, and $C$ is bounded, then $A \subset B.$
Is it essential that $A$ be convex? 
 A: No. Actually, the original version of the lemma does not require the convexity of A.
Reference: Hans Rådström (note spelling): An embedding theorem for spaces of convex sets.
Proc. Amer. Math. Soc. 3 (1952) 165–169, MR0045938. doi:10.2307/2032477.
A: Alex, your idea of using a separation theorem can be turned into a full proof as follows:
As you said, $A \not\subset B$ iff there is a $a \in A$ and vector $x$ such that $\sup_{b \in B} \langle b,x \rangle < \langle a,x \rangle$. But then,
\begin{align}
\sup_{z \in B + C} \langle z, x \rangle &= \sup_{b \in B, c \in C} (\langle b, x \rangle + \langle c, x \rangle ) \\\\
&= \sup_{b \in B} \langle b, x \rangle + \sup_{c \in C} \langle c, x \rangle \\\\
&< \langle a, x \rangle + \sup_{c \in C} \langle c, x \rangle \\\\
&= \sup_{z \in a + C} \langle z, x \rangle \\\\
&\leq \sup_{z \in B + C} \langle z, x \rangle,
\end{align}
a contradiction.
A: Hmm, I have an idea for a proof which works even when $A$ isn't convex:
$A \not\subset B$ iff there is a $a \in A$ and vector $x$ such that $\langle a, x\rangle >0$ and $\langle b, x \rangle \leq 0$ for all $b \in B$ (since $B$ is a closed convex set). Let $b \in B$ and $c \in C$ maximize $\langle b + c, x \rangle$, then $\langle a + c, x \rangle > \langle b + c, x \rangle$, so there is a point in $A + C$ that isn't in $B+C$.
This implies that if $A \not\subset B$ then $A + C \not\subset B+C$. Of course, there aren't really $b,c$ which maximize that quantity above, but I believe you could make this rigorous using approximation arguments.
