Continuous separation when one convex set is open

First work over the reals and let $D$ be nonempty open convex with $z\mathrm{\notin }D$. Fix $d_0\in D$ and put $U=D-d_0$ and $v=z-d_0$. Then $U$ is an open convex zero-neighborhood, $v\mathrm{\ne }0$, and $p_U(v)\ge 1$ because $v\mathrm{\notin }U$. The set $M=\{tv:t\in \mathbb{R}\}$ is a real subspace: sums and real multiples remain on the line. If $tv=sv$ then multiplying $(t-s)v=0$ by $(t-s{)}^{-1}$ when $t\mathrm{\ne }s$ would give $v=0$, so the coefficient is unique. Thus $h(tv)=t$ is well-defined and real-linear.

For $t\ge 0$, $h(tv)=t\le t{p}_U(v)=p_U(tv)$; for $t<0$, $h(tv)=t<0\le p_U(tv)$. All hypotheses of the relative extension theorem are now met. Apply HB once to obtain real-linear $H:X\to \mathbb{R}$ extending $h$ with $-p_U(-x)\le H(x)\le p_U(x)$. In particular $H(v)=1$. This is the only non-ZF input in the proof.

On the open zero-neighborhood $U\cap (-U)$ both gauges $p_U(x),p_U(-x)$ are less than one. Hence $\mathrm{\mid }H(x)\mathrm{\mid }<1$ there, so $H$ is continuous by the modulus-bound criterion. For $d\in D$, $H(d)-H(d_0)=H(d-d_0)\le p_U(d-d_0)<1=H(z)-H(d_0)$. Therefore $H(d)<H(z)$, and $H\mathrm{\ne }0$ since $H(v)=1$.

For the given real $A,B$, take $D=A-B$ and $z=0$. This is open, since it is the union of the translates $A-b$ for $b\in B$; it is convex by distributing each real convex combination through the difference. It is nonempty and excludes zero by disjointness. The preceding construction therefore produces a nonzero continuous real-linear $H$ with $H(a-b)<0$, or $H(a)<H(b)$ for every $a,b$. It also produces a vector $v$ with $H(v)=1$.

The set $-H(A)$ is nonempty and bounded below by $-H(b_0)$ for any fixed $b_0\in B$. Let $\alpha =-\inf (-H(A))$. By reversing the defining lower-bound inequalities, $\alpha$ is the least upper bound of $H(A)$. Every $H(b)$ is an upper bound, so $\alpha \le H(b)$. For $a\in A$, openness and continuity of $s\mapsto a+sv$ give a positive $s$ with $a+sv\in A$; hence $H(a)<H(a)+s\le \alpha$. If $B$ is open and $H(b)=\alpha$, a small $s>0$ with $b-sv\in B$ would give $\alpha \le H(b-sv)=\alpha -s$, an impossibility. Thus both inequalities are strict when both sets are open.

For complex $X$, restrict scalars to $\mathbb{R}$. The scalar inclusion $\mathbb{R}\to \mathbb{C}$ is continuous since it preserves distance, so the restricted scalar action is jointly continuous. Convexity and openness are unchanged. Apply the real construction to obtain $H$ and $\alpha$. Set $f(x)=H(x)-iH(ix)$. It is additive and real-homogeneous, and $f(ix)=H(ix)-iH(-x)=H(ix)+iH(x)=if(x)$. For $\lambda =a+ib$, additivity and real homogeneity therefore give $f(\lambda x)=af(x)+bf(ix)=\lambda f(x)$. Both $H$ and $x\mapsto H(ix)$ are continuous; scalar addition and multiplication are continuous, so $f$ is continuous. Its real part is $H$, so it is nonzero and obeys the same inequalities. This proves the complex case as well, with the same single HB application and no assumption of AC.