Your argument is correct. You actually proved that $A_{j_1, \dots, j_n}$ has diameter at most $\epsilon$ (which of course implies it's "contained in an $\varepsilon$-ball"). Personally I would stick with diameter formulation, because phrasing the result in terms of $\varepsilon$-ball makes the reader wonder what the center of that ball might be (which we don't know and don't care about; in fact $A_{j_1, \dots, j_n}$ may well be empty).
I edited this post, because I misunderstood the problem. However the solution is almost the same. Please check it.
Let us prove that there is a vector space $V$ and two convex subsets of $V$, $A$ and $B$, such that the only functional $f\in V^*$ with the following property is the zero functional.
The property is for every $a\in A$ and $b\in B$, there is $\alpha\in \mathbb{R}$ such that $f(a)\leq\alpha\leq f(b)$.
Consider the vector space of real sequences $\mathbb{R}^{\mathbb{N}}$.
Let $A$ be the subspace of $\mathbb{R}^{\mathbb{N}}$ formed by sequences with finitely many non zero coordinates. Notice that $A$ is convex since it is a vector space.
Let $B$ be the subset of $\mathbb{R}^{\mathbb{N}}$ formed by sequences of non negative numbers converging to $0$ with infinitely many non zero coordinates. Notice that $A\cap B=\emptyset$ and any convex combination of elements in $B$ still belongs to $B$, thus $B$ is also convex.
Now, let $V=\text{span } A\cup B$.
Let $f\in V^*$ be any linear functional such that $f(a)\leq\alpha\in\mathbb{R}$, for every $a\in A$.
Since $A$ is a subspace, the only possibility is $f(a)=0$, for every $a\in A$.
Suppose that $f(b)\geq 0$ for every $b\in B$.
Let $(b_n)_{n\in\mathbb{N}}\in B$ and notice that $(\sqrt{b_n})_{n\in\mathbb{N}}\in B$.
Let $k>0$.
Notice that $\dfrac{\sqrt{b_n}}{k}-b_n=\sqrt{b_n}(\dfrac{1}{k}-\sqrt{b_n})$ and since $(\sqrt{b_n})_{n\in\mathbb{N}}$ converges to $0$, there is $N\in\mathbb{N}$ such that $\dfrac{1}{k}-\sqrt{b_n}> 0$ for $n>N$. Therefore, $\dfrac{\sqrt{b_n}}{k}-b_n\geq 0$ for $n>N$.
Define $c_n=0$ for $n\leq N$ and $c_n=\dfrac{\sqrt{b_n}}{k}-b_n$ for $n>N$. Thus, $(c_n)_{n\in\mathbb{N}}\in B$.
Next, define $d_n=\dfrac{\sqrt{b_n}}{k}-b_n$ for $n\leq N$ and $d_n=0$ for $n>N$. Thus, $(d_n)_{n\in\mathbb{N}}\in A$.
Therefore $(\dfrac{\sqrt{b_n}}{k})_{n\in\mathbb{N}}-(b_n)_{n\in\mathbb{N}}=(c_n)_{n\in\mathbb{N}}+(d_n)_{n\in\mathbb{N}}$.
Now, $f((\dfrac{\sqrt{b_n}}{k}))-f((b_n))=f((c_n))+f((d_n))=f((c_n))\geq 0$ then $$\frac{1}{k}f((\sqrt{b_n}))\geq f((b_n))\geq 0.$$
Since $k$ is arbitrary. This inequality implies that $f((b_n)_{n\in\mathbb{N}})=0$. Now, $(b_n)_{n\in\mathbb{N}}$ is any element of $B$. Thus, $f(b)=0$ for every $b\in B$.
Since $f\in V^*$ and $V=\text{span }A\cup B$ then $f=0$.
Best Answer
Probably the easiest way to prove this is using sequences. Suppose $(c_n)$ is a sequence in $C$ converging to some point $x\in E$. For each $n$, we can write $c_n=a_n-b_n$ where $a_n\in A$ and $b_n\in B$. Since $B$ is compact, we can pass to a subsequence and assume that $(b_n)$ converges to some $b\in B$. But now since $c_n\to x$ and $b_n\to b$, $a_n=b_n+c_n$ must converge to $b+x$. Since $A$ is closed, this means $b+x\in A$. Thus $x=(b+x)-b\in C$, as desired.