convexity - Radstrom cancellation only for two convex sets?

Alex, your idea of using a separation theorem can be turned into a full proof as follows:

As you said, $A notsubset B$ 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.

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