Counterexample of 2nd geometric form of Hahn-Banach theorem

Question

I recall the 2nd geometric form of Hahn-Banach theorem :

Let $E$ be a real locally-convex topological vector space. Let $A$ and $B$ be two non-empty disjoint and convex subsets of $E$, $A$ being closed and $B$ being compact. Then there are a bounded linear form $f:E\rightarrow\mathbb{R}$ and two constants $a\in\mathbb{R}$, $\varepsilon>0$ such that \begin{cases} \forall x\in A,\ f(x)\leq a-\varepsilon\\\forall y\in B,\ f(x) \geq a+\varepsilon \end{cases}

Is there a "simple" example of a non locally-convex topological vector space that provides a counterexample to this theorem ? The only example of a non locally-convex topological vector space i know comes from the metric space $(L^p([0,1]),d)$ for $0<p<1$ where the metric $d$ is defined as such

$$\forall g,h\in L^p([0,1]),\quad d(g,h)=\int_0^1 |g(t)-h(t)|^p\,dt$$

but its only convex subset is $L^p([0,1])$ itself, which can't provide such a counterexample.

Thanks in advance !

Answer 1

(from mihaild) Take two different points in $L^{1/2}$. Since the only continuous linear form over $L^{1/2}$ is $t\mapsto 0$, we can't separate these singletons.