Coordinate functionals give an explicit uniform separating gap

Pointwise operations give $f(x+y)=f(x)+f(y)$ and $f(cx)=cf(x)$, so the projection $f$ is scalar-linear; it is continuous. Thus $u=\mathrm{Re}f$ is continuous and real-linear. The rays $(-\mathrm{\infty },a]$ and $[b,\mathrm{\infty })$ are closed: their complements are unions of open intervals. Their inverse images under $u$ are therefore closed, since inverse images of their open complements are open. They are convex because real-linear maps preserve real convex combinations and each ray is convex.

The constant functions with values $a$ and $b$ belong to $A$ and $B$, respectively, so both sets are nonempty. Their intersection is empty since $b>a$. Set $\alpha =(a+b)\mathrm{/}2$ and $\epsilon =(b-a)\mathrm{/}2>0$. Then for every $x\in A,y\in B$, $\mathrm{Re}f(x)\le a=\alpha -\epsilon <\alpha +\epsilon =b\le \mathrm{Re}f(y)\mathrm{.}$ The constant-one function has $f(1)=1$, so $f$ is nonzero.

For distinct $x,y$, fix one index $i$ with $z=x_i-y_i\mathrm{\ne }0$. Over $\mathbb{C}$ put $c=\bar{z}\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$; then $\mathrm{\mid }c\mathrm{\mid }=1$ and $cz=\mathrm{\mid }z\mathrm{\mid }>0$. Over $\mathbb{R}$ put $c=z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$, which gives the same identities. Thus $g(w)=c{w}_i$ is continuous and scalar-linear, and $\mathrm{Re}g(x)-\mathrm{Re}g(y)=\mathrm{\mid }z\mathrm{\mid }>0$. The index and unit scalar are chosen for this one supplied pair; there is no simultaneous choice. If $I$ is empty there is no $i_0$ and no pair of distinct functions, so the respective hypotheses do not arise.