= Solution
Let $p$ be the <Minkowski functional> of $E$:
$$
p(v)=\inf\{t>0:v\in tE\}.
$$
The contained ball makes $E$ absorbing, so $p$ is finite, nonnegative and satisfies $p(v)\leq\|v\|/\epsilon$. It is positively homogeneous. To check subadditivity, if $v\in sE$ and $w\in tE$, the defining property of a <convex set> gives $(v+w)/(s+t)\in E$, whence $p(v+w)\leq s+t$; let $s,t$ decrease to the respective infima. Thus $p$ is a <sublinear functional>. We have $p(e)\leq1$ for every $e\in E$. Also $p(x)\geq1$: otherwise $x\in tE$ for some $0<t<1$, which would put $x$ in $E$ because $E$ is a <convex set> and $0\in E$.
We prove the dominated form of the <Hahn-Banach theorem> needed here. Suppose $g$ is linear on a subspace $M$ and $g\leq p$ there. To extend it to $M+\mathbb Rv$ with $v\notin M$, choose the value $c$ at $v$ between
$$
\sup_{m\in M}\{g(m)-p(m-v)\}
\quad\text{and}\quad
\inf_{n\in M}\{p(n+v)-g(n)\}.
$$
Every lower candidate is at most every upper candidate, because
$$
g(m)+g(n)=g(m+n)\leq p(m+n)\leq p(m-v)+p(n+v).
$$
Taking $m=n=0$ among the candidates shows the two endpoints are finite: the lower supremum is at least $-p(-v)$ and at most $p(v)$, and the upper infimum lies between them and $p(v)$. Define $\widetilde g(m+tv)=g(m)+tc$. For $t>0$, the upper inequality applied to $m/t$ gives domination by $p(m+tv)$. For $t<0$, apply the lower inequality to $m/(-t)$ and multiply by $-t$. At $t=0$ domination already holds. Hence this is a dominated one-dimensional extension.
Order all dominated extensions by inclusion of their domains. Chain unions preserve linearity and domination, so <Zorn's lemma> gives a maximal extension. The one-dimensional argument shows its domain is the whole vector space. This proves the requisite dominated <Hahn-Banach theorem>, rather than invoking an unproved separation form.
On the line $\mathbb Rx$, start with $g(tx)=tp(x)$. For $t\geq0$ this equals $p(tx)$; for $t<0$ it is nonpositive and therefore at most $p(tx)$. Extend it to $T\leq p$ on $V$. Then
$$
Tx=p(x)\geq1,\qquad Te\leq p(e)\leq1\quad(e\in E).
$$
Applying domination to $v$ and $-v$ also gives
$$
|Tv|\leq\frac{\|v\|}{\epsilon}.
$$
Thus $T$ is a <continuous linear functional>, and
$$
\boxed{Tx\geq1\geq Te\quad\text{for every }e\in E.}
$$
This proves <separation from an absorbing convex set> with no closedness assumption on $E$. The proof also gives the norm-preserving real extension theorem used earlier: take $p(v)=C\|v\|$ when the original functional has <norm> at most $C$.
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