A closed convex set is an intersection of closed half-spaces

Since every closed affine half-space in the family contains $C$, their intersection contains $C$.

If $x\mathrm{\notin }C$, choose $r>0$ with $B(x,r)\cap C=\mathrm{\varnothing }$. The set $C+B(0,r\mathrm{/}2)$ is open and convex and still omits $x$; [F1], using AC through its open-convex point-separation supplier, gives a nonzero continuous $f$ with $\mathrm{Re}f(c+b)<\mathrm{Re}f(x)$ for every $c\in C$ and $\mathrm{\parallel }b\mathrm{\parallel }<r\mathrm{/}2$.

Taking the supremum over $\mathrm{\parallel }b\mathrm{\parallel }<r\mathrm{/}2$ in step 1.2 gives $\mathrm{Re}f(c)+(r\mathrm{/}2)\mathrm{\parallel }f\mathrm{\parallel }\le \mathrm{Re}f(x)$ for every $c\in C$. Since $f\mathrm{\ne }0$, the closed half-space $H_x=\{z:\mathrm{Re}f(z)\le \mathrm{Re}f(x)-(r\mathrm{/}4)\mathrm{\parallel }f\mathrm{\parallel }\}$ contains $C$ and excludes $x$. Thus every exterior point is excluded from the intersection, proving equality.