= Solution
We first prove the algebraic extension step needed for <Hahn-Banach separation theorem>. Suppose $p$ is a finite <sublinear functional> on a real <vector space> and $g$ is a <linear functional> on a <vector subspace> $W$, with $g\leq p$ there. For $v\notin W$, choose a real number $c$ between
$$
\sup_{w\in W}\{g(w)-p(w-v)\}
\quad\hbox{and}\quad
\inf_{w\in W}\{p(w+v)-g(w)\}.
$$
This interval is nonempty: for any $w,z\in W$,
$$
g(w)+g(z)=g(w+z)\leq p(w+z)\leq p(w-v)+p(z+v).
$$
Taking $w=0$ or $z=0$ also bounds the endpoints by finite numbers. Define $\widetilde g(w+tv)=g(w)+tc$. For $t>0$, the upper bound for $c$, applied to $w/t$, proves $\widetilde g(w+tv)\leq p(w+tv)$. For $t<0$, writing $t=-s$ and using the lower bound at $w/s$ proves the same inequality. The case $t=0$ is the original domination. Thus a dominated <linear functional> extends over one further direction. Order all dominated extensions by extension of their domains. A chain has its union as an upper bound, so <Zorn's lemma> supplies a maximal extension. The one-direction argument shows its domain is the whole <vector space>. This proves the required algebraic <Hahn-Banach theorem>, rather than assuming the separation result.
Choose $a\in A$, and put $C=A-a$. This is a <radially open convex set> containing zero. It is absorbing: along each line through zero, sufficiently small multiples of any vector lie in $C$. Its <Minkowski functional>
$$
p(x)=\inf\{t>0:x\in tC\}
$$
is finite and nonnegative. Positive homogeneity follows by rescaling $t$. If $x\in sC$ and $y\in tC$, convexity gives $x+y\in(s+t)C$, proving <subadditivity> by taking infima. Moreover,
$$
C=\{x:p(x)<1\}.
$$
Indeed, $p(x)<1$ gives $x\in sC$ for some $s<1$, hence $x\in C$ by convexity and $0\in C$. Conversely, radial openness at $x\in C$ permits $(1+\epsilon)x\in C$ for some $\epsilon>0$, giving $p(x)<1$.
Since $A\cap U=\varnothing$, $a\notin U$ and $u-a\notin C$ for every $u\in U$. Define a <linear functional> on $U+\mathbb Ra$ by
$$
g(u+ta)=-t.
$$
It is well defined because $a\notin U$. For $t<0$, write $s=-t$; then $p(u+ta)=s\,p(u/s-a)\geq s=g(u+ta)$. For $t\geq0$, domination follows from $g(u+ta)\leq0\leq p(u+ta)$. The extension step therefore gives $L:V\to\mathbb R$ with $L\leq p$, $L|_U=0$ and $L(a)=-1$. If $x\in A$, then
$$
L(x)+1=L(x-a)\leq p(x-a)<1,
$$
so $L(x)<0$. Consequently
$$
\boxed{H=\ker L\supseteq U,\qquad H\cap A=\varnothing.}
$$
Because $L(a)\ne0$, this <kernel of a linear map> has codimension one and is a <hyperplane>. This establishes the <geometric Hahn-Banach theorem for radially open sets> without assuming a norm or a topology on $V$.
For the <convex function>, use its strict <epigraph> in $V\times\mathbb R$. It is nonempty and convex, and does not contain $(0,0)$. It is also radially open: on every affine line, a finite <convex function> is continuous. To justify this last fact, convexity orders secant slopes; on a smaller interval the slopes are bounded above and below by slopes to two fixed exterior endpoints, giving a local Lipschitz bound. Thus $t\mapsto f(x+ty)-\beta-ts$ is continuous, and a strict negative value persists near $t=0$.
Apply the separation theorem with the zero subspace. A nonzero separating <linear functional> has form $L(x,\beta)=g(x)+c\beta$. Its values on the convex <epigraph> cannot have both signs, since a line segment would then meet its kernel. Choose the sign so that $L>0$ on the <epigraph>. At $x=0$ and $\beta>0$ this gives $c>0$. Letting $\beta$ decrease to $f(x)$ yields $g(x)+cf(x)\geq0$. Therefore
$$
\boxed{l(x)=-g(x)/c\leq f(x)\quad\hbox{for every }x\in V.}
$$
The resulting <linear functional> need not be nonzero: for example, the zero functional is the only linear minorant of $f(x)=x^2$ on $\mathbb R$.
Now suppose the norm and continuity hypotheses hold. Domination at $x$ and $-x$ gives
$$
-f(-x)\leq l(x)\leq f(x).
$$
If $x_n\in\ker l$ and $x_n\to x$ in norm, apply this inequality to $x-x_n$. Since $l(x-x_n)=l(x)$ and both bounding values tend to $f(0)=0$, we obtain $l(x)=0$. The <kernel of a linear map> is therefore closed.
Finally, a <linear functional with closed kernel> on a <normed vector space> is continuous. If $l=0$, this is immediate. Otherwise choose $v$ with $l(v)=1$ and put $M=\ker l$. Closedness gives $d=\operatorname{dist}(v,M)>0$. For $l(x)\ne0$, $x/l(x)-v\in M$, so $d\leq\|x\|/|l(x)|$. The same resulting bound holds when $l(x)=0$:
$$
\boxed{|l(x)|\leq d^{-1}\|x\|.}
$$
This proves continuity and explains exactly how closedness of the kernel is used.
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