The statement you want to prove is the classic first version of the Hahn-Banach Separation Theorem .
Usually one has available the Hahn-Banach Theorem, in the form
If $M\subset X$ is a subspace, $\varphi:M\to\mathbb R$ is linear, and $\mu$ is a real seminorm on $X$ (subadditive, $\mu(\lambda x)=\lambda\,\mu(x)$ for $\lambda\geq0$) with $|\varphi(x)|\leq\mu(x)$ for all $x\in M$, then there exists $\tilde\varphi:X\to\mathbb R$, linear, with $\tilde\varphi|_M=\varphi$ and $|\tilde\varphi(x)|\leq\mu(x)$ for all $x$.
Since the interior of a convex set is convex, we may assume that $E$ is open (as the inequality will hold for any limit point). Fix $e'\in E$, $f'\in F$; The condition $E\cap F=\varnothing$ guarantees $e'\ne f'$. Then the set $Z=E-F+f'-e'$ is convex, open and $0\in E-F+f'-e'$. If $\mu$ is its Minkowski Functional, the conditions on $Z$ (open convex neighbourhood of $0$) guarantee that
$$\tag1
Z=\{x\in X:\ \mu(x)<1\}.
$$
On the subspace $M=\mathbb R(f'-e')$, define $\varphi:M\to\mathbb R$ by $\varphi(\lambda(f'-e'))=\lambda$. Then $\varphi$ is linear. As $f'-e'\not\in Z$, we have $\mu(f'-e')\geq1$. So
$$
|\varphi(\lambda(f'-e'))|=|\lambda|\leq|\lambda|\,\mu(f'-e')=\mu(\lambda(f'-e')).
$$
That is, $|\varphi(x)|\leq\mu(x)$ for all $x\in M$. By the Hahn-Banach Theorem, there exists $\tilde\varphi:X\to\mathbb R$, linear, with $|\tilde\varphi(x)|\leq\mu(x)$ for all $x\in X$.
Now, for any $e\in E$, $f\in F$,
$$
\tilde\varphi(e)-\tilde\varphi(f)+1=\tilde\varphi(e-f+f'-e)<1
$$
by $(1)$. Thus
$$\tag2
\tilde\varphi(e)<\tilde\varphi(f),\qquad e\in E,\ f\in F.
$$
Let $c=\sup\{\tilde\varphi(e):\ e\in E\}$. Then $c\leq\tilde\varphi(f)$ for all $f\in F$. Using that $F$ is a subspace, we get $c\leq t\tilde\varphi(f)$ for all $t\in\mathbb R$. This forces $\tilde\varphi(f)=0$ for all $f\in F$. Thus
$$\tag3
\tilde\varphi(f)=0,\ \tilde\varphi(e)\leq0,\qquad f\in F,\ e\in E.
$$
Finally, limit points of $E$ are also limit points of $Z$, and so they satisfy $\mu(x)\leq1$ as the Minkowski Functional is continuous. So the argument above still works, just with $\leq$ instead of $<$.
Best Answer
Check that
$$p \colon x \mapsto \operatorname{dist}(x,U)$$
is a sublinear functional. (It's a seminorm, actually.)
Then apply the Hahn-Banach extension theorem to $\tilde{p} = \frac{1}{\operatorname{dist}(z,U)}\cdot p$, and the linear map $t\cdot z \mapsto t$ from the one-dimensional subspace spanned by $z$ to the scalar field.
Alternatively, let $W = z - \overline{U}$, and $C = B_{\operatorname{dist}(z,U)}(0)$. Then $W$ and $C$ are disjoint convex sets, and $C$ is open and balanced. By one of the Hahn-Banach separation theorems, there is a continuous linear functional $\psi_0 \colon V \to \mathbb{R}$ with $\lvert \psi_0(x)\rvert < 1$ for all $x\in C$ and $\psi_0(y) = 1$ for all $y \in W$. These conditions show $\lVert \psi_0\rVert = \operatorname{dist}(z,U)^{-1}$ and $\psi_0(z) = 1$. If the scalar field is $\mathbb{R}$, let $\psi = \psi_0$, if the scalar field is $\mathbb{C}$, let $\psi$ the continuous linear form with real part $\psi_0$, that is, $\psi(x) = \psi_0(x) - i \psi_0(ix)$ for all $x\in V$.