Cancellation law for Minkowski sums

Let $(X,\|\cdot\|)$ be a Banach space and $A,B,C\subset X$ closed bounded non-empty convex subsets. Let $+$ denote the Minkowski symbol for addition.

Does the $+$ satisfy: $$A+C\subset B+C\implies A\subset B$$

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It does. Suppose $a\in A$ with $a\notin B$. Since $B$ is closed and not empty, there is $b\in B$ with minimal distance from $A$. Project everything onto $a-b$, and denote the projected objects by primes. Since the sets are closed and convex, their projections are closed intervals. $B'$ is all on one side of $a'$, since the line segment from $b$ to a point whose projection is on the other side of $a'$ would contain points closer to $a$ than $b$. Thus $a'+C'\not\subset B'+C'$, hence $a+C\not\subset B+C$ and thus $A+C\not\subset B+C$.

[Edit:]

As t.b. pointed out, this doesn't work in general, but as D. Thomine pointed out, it can be fixed using the Hahn–Banach theorem. Again, suppose $a\in A$ with $a\notin B$. Since $\{a\}$ is convex and compact and $B$ is convex and closed, there is a continuous linear map $\lambda:X\to\mathbb R$ strictly separating the two, so there is $s\in\mathbb R$ such that $\lambda(a)\lt s\lt\lambda(b)$ for all $b\in B$. Since $C$ is bounded and $\lambda$ is continuous, $\lambda(C)$ is bounded. Thus $\lambda(a)+\lambda(C)\not\subset\lambda(B)+\lambda(C)$, hence $a+C\not\subset B+C$ and thus $A+C\not\subset B+C$.

Note that only the boundedness of $C$, not that of $A$ or $B$ has been used.

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How do you get that $b \in B$ with minimal distance from $A$ and how exactly do you project? I think this argument works fine if you assume the space to be uniformly convex (hence +1) but I don't see how to do this in full generality. Maybe I'm missing something simple here. – t.b. Jul 25 '12 at 13:46
@t.b. : in this case I think we can work with Hanh-Banach. Find an hyperplane which strictly separates $a$ and $B$; its direction will play the role of $a-b$ in Joriki's argument. – D. Thomine Jul 25 '12 at 13:57
@D.Thomine Yes, that works. Thanks! – t.b. Jul 25 '12 at 14:06
There is now a follow-up question here where it is claimed that the present question isn't properly answered. Maybe you want to add a line about the fix suggested by D. Thomine? – t.b. Aug 14 '12 at 20:24
@t.b.: Thanks for notifying me -- sorry I couldn't respond immediately; I travelled to Palestine the day you wrote and it took me a while to get net access. I've completed the proof now. I had also failed to point out where I used the boundedness of $C$ in the first version; perhaps you can check that I got the details right this time? – joriki Aug 18 '12 at 4:48