Prove that if $A, B $ are convex , $B$ is closed, $C$ is bounded and $A+C \subset B+C$ then $A\subset B$ Given the sets $A,B,C \in \mathbb{R}^n$ such that:
$$A+C \subset B+C$$
Show that if $A,B$ are convex, $B$ is closed and $C$ is bounded then $A\subset B$.
I kind of understand the geometrical interpretation of the problem but I fail to see where to start, too many properties for me to handle :( 

It seems that the following  plan will work, and not only for the case of $\mathbb{R}^n$, but for a general case of a locally convex space $X$, satisfying the Hahn-Banach Theorem.
The convexity of $A$ is not needed, because the question condition implies $a+C\subset B+C$  for each point $a\in A$. So it suffices to consider only a case when $A$ is one-point set. Without loss of generality, we may suppose that $A={0}$. Then we should prove that $C\subset B+C$ implies $0\in B$. Suppose the opposite. Since $B$ is closed, there exist a real number $\varepsilon>0$ and a linear functional $f$ on $X$ such that $f(0)=0$ and $f|B>\varepsilon$. Since $C$ is bounded, the value $c=\inf_{x\in C} f(x)<\infty$. Choose a point $x\in C$ such that $f(x)<c+\varepsilon/2$. Since $f(y+z)=f(y)+f(z)>\varepsilon+c$ for each $y\in B$ and $z\in C$, we see that $x\not\in B+C$, a contradiction. 
