I am trying to prove the following.
Question. Suppose $A,B,C\subset E^n$ and that $$A+C \subset B+C.$$
If $A,B$ are both convex, $B$ is closed, $C$ is bounded, show that $A\subset B$.
So far I have done the following. Where am I going wrong?
Proof. Let $a\in A$. Then, $$a+c_0\in A+C,\ \forall c_0\in C$$ $$ \Rightarrow a+c_0 \in B+C, \forall c_0\in C$$ $$ \Rightarrow a+c_0 = b+c_1,\ \text{for some}\ b\in B,c_1\in C.$$
For every line $\lambda a_0 + (1-\lambda)a_1 + c_0,\ 0\leq \lambda \leq 1,$ that contain $a$, there must exist a closed line $\lambda b_0 + (1-\lambda)b_1 + c_1, 0\leq \lambda \leq 1,$ that contains $b$ and $\lambda a_0 + (1-\lambda)a_1 + c_0$. Now, since $C$ is bounded $a$ is contained in $\lambda b_0 + (1-\lambda)b_1\in B$.
Suppose on the contrary that $A \not \subset B$. Then there exists $a \in A$ such that $a \not \in B$. Since $B$ is closed and convex and so is $\{a\}$, from some version of the geometric Hahn-Banach theorem, there are a real number $r$ and a nonzero linear map $l : E^n \to \mathbb{R}$ such that $l(a) > r$ and $l(B) \le r$. Set $\epsilon = l(a)-r > 0$.
Since $C$ is bounded, since $l$ is linear and since we work in finite dimension $n$, the number $M := \sup_{c \in C} l(c)$ is finite. Observe that for $x \in B+C$, $l(x) \le r + M$. Consider $c_0 \in C$ such that $l(c_0) > M - (\epsilon/2)$. Then by the linearity of $l$, $$l(a+c_0) = l(a) + l(c_0) > r + \epsilon + M - (\epsilon/2) = r + M + (\epsilon/2) > r+M \, .$$ It follows that $a + c_0 \not \in B+C$, which contradicts the assumption that $A+C \subset B+C$. Notice that we did not use the convexity of $A$.
The geometric Hahn-Banach theorem used above is a rather accessible statement in finite dimensions. Let $B \subset E^n$ be a closed convex set and $\{a\} \in E^n \setminus B$.
Since convexity is a property invariant under changes of equivalent norms, and since all norms on a finite dimensional space are equivalent, we can without loss of generality consider the strictly convex norm $\|-\|$ on $E^n$ associated to the 'usual' Euclidean scalar product on $\mathbb{R}^n$.
Let $d = d(a, B) = \mathrm{inf}_{b \in B} \, d(a,b)$ where $d(x,y) := \|y-x\|$; we claim that this infimum is achieved. Indeed, for $n \in \mathbb{N}$, let $b_n \in B$ be such that $d(a,b_n) \le d + (1/n)$. (Observe interestingly that this relies on the countable axiom of choice, but this is a weaker axiom than the usual axiom of choice and it is somewhat more 'intuitive' than Zorn's lemma...) Since the sequence $\{b_n\}$ lies inside the bounded closed (hence compact) set $B \cap \bar{B}(a, 1)$, a subsequence of it converges say towards $b_0 \in B$. Of course, $d = d(a,b_0)$.
We claim that $b_0$ is the unique element in $b \in B$ such that $d = d(a,b)$. Assuming otherwise, let $b_1 \in B \setminus \{b_0\}$ satisfy $d = d(a,b_1)$. Since $B$ is convex, the segment $b_{\lambda} = \lambda b_0 + (1- \lambda) b_1$ with $\lambda \in [0,1]$ lies in $B$. Hence for $\lambda \in (0,1)$ $$ d \le d(a, b_{\lambda}) = \| b_{\lambda} - a \| = \| \lambda (b_0 - a) + (1- \lambda)(b_1 - a) \| < \lambda d(a,b_0) + (1- \lambda) d(a,b_1) = d$$ where the strict inequality holds since the euclidean norm is strictly convex. This is a contradiction, hence $b_0$ is unique.
Let $v = a-b_0$ and $w = (b_0 + a)/2$. Consider the linear map $l : E^n \to \mathbb{R} : x \mapsto x \cdot v$ and set $r := l(w)$. Then $l(a) > r$ and $l(b_0) < r$. Moreover $l(B) < r$, in fact $\mathrm{max}_{b \in B} l(b) = l(b_0)$; this follows from the fact that the hyperplane $[x:l(x) = l(b_0)]$ is the tangent plane at $b_0$ to the sphere $\partial B(a, d)$. Indeed, for $x \in \bar{B}(a,d)$, $l(x) \ge l(b_0)$ with equality if and only if $x=b_0$. If there were to exist $b_1 \in B$ such that $l(b_1) > l(b_0)$, a part of the segment joining $b_0$ to $b_1$ would lie strictly inside $\bar{B}(a,d)$ hence contradicting the definition of $b_0$.