Can any two disjoint nonempty convex sets in a vector space be separated by a hyperplane?

Let $V$ be a normed vector space over $\mathbb{R}$, and let $A$ and $B$ be two disjoint nonempty convex subsets of $V$. A geometric form of Hahn-Banach Theorem states that $A$ and $B$ can be separated by a closed hyperplane (i.e. there is $f \in V^\ast$ and $\alpha \in \mathbb{R}$ such that $f(a) \le \alpha, \forall a \in A$ and $\alpha \le f(b), \forall b \in B$) if either $A$ or $B$ is open, or $A$ is closed and $B$ is compact. (This statement is not in the full generality.) There are examples of two disjoint nonempty convex sets which cannot be separated by a closed hyperplane.(These convex sets don't satisfy the condition of the previous statement.)

My question is: If the separating hyperplane need not be closed, can any pair of disjoint nonempty convex sets be separated by a hyperplane? More precisely, for any vector space $V$ over $\mathbb{R}$ and two disjoint nonempty convex subsets $A$ and $B$ of $V$, does there exist a linear functional $f:V\to\mathbb{R}$ and a real number $\alpha\in\mathbb{R}$ such that $f(a) \le \alpha, \forall a \in A$ and $\alpha \le f(b), \forall b \in B$?

This question doesn't involve any topological concepts. For the finite dimensional case, it is known that the separation is possible. What would happen if the underlying space is infinite dimensional?

I edited this post, because I misunderstood the problem. However the solution is almost the same. Please check it.

Let us prove that there is a vector space $V$ and two convex subsets of $V$, $A$ and $B$, such that the only functional $f\in V^*$ with the following property is the zero functional. The property is for every $a\in A$ and $b\in B$, there is $\alpha\in \mathbb{R}$ such that $f(a)\leq\alpha\leq f(b)$.

Consider the vector space of real sequences $\mathbb{R}^{\mathbb{N}}$.

Let $A$ be the subspace of $\mathbb{R}^{\mathbb{N}}$ formed by sequences with finitely many non zero coordinates. Notice that $A$ is convex since it is a vector space.

Let $B$ be the subset of $\mathbb{R}^{\mathbb{N}}$ formed by sequences of non negative numbers converging to $0$ with infinitely many non zero coordinates. Notice that $A\cap B=\emptyset$ and any convex combination of elements in $B$ still belongs to $B$, thus $B$ is also convex.

Now, let $V=\text{span } A\cup B$. Let $f\in V^*$ be any linear functional such that $f(a)\leq\alpha\in\mathbb{R}$, for every $a\in A$. Since $A$ is a subspace, the only possibility is $f(a)=0$, for every $a\in A$.

Suppose that $f(b)\geq 0$ for every $b\in B$.

Let $(b_n)_{n\in\mathbb{N}}\in B$ and notice that $(\sqrt{b_n})_{n\in\mathbb{N}}\in B$. Let $k>0$.

Notice that $\dfrac{\sqrt{b_n}}{k}-b_n=\sqrt{b_n}(\dfrac{1}{k}-\sqrt{b_n})$ and since $(\sqrt{b_n})_{n\in\mathbb{N}}$ converges to $0$, there is $N\in\mathbb{N}$ such that $\dfrac{1}{k}-\sqrt{b_n}> 0$ for $n>N$. Therefore, $\dfrac{\sqrt{b_n}}{k}-b_n\geq 0$ for $n>N$.

Define $c_n=0$ for $n\leq N$ and $c_n=\dfrac{\sqrt{b_n}}{k}-b_n$ for $n>N$. Thus, $(c_n)_{n\in\mathbb{N}}\in B$.

Next, define $d_n=\dfrac{\sqrt{b_n}}{k}-b_n$ for $n\leq N$ and $d_n=0$ for $n>N$. Thus, $(d_n)_{n\in\mathbb{N}}\in A$.

Therefore $(\dfrac{\sqrt{b_n}}{k})_{n\in\mathbb{N}}-(b_n)_{n\in\mathbb{N}}=(c_n)_{n\in\mathbb{N}}+(d_n)_{n\in\mathbb{N}}$.

Now, $f((\dfrac{\sqrt{b_n}}{k}))-f((b_n))=f((c_n))+f((d_n))=f((c_n))\geq 0$ then $$\frac{1}{k}f((\sqrt{b_n}))\geq f((b_n))\geq 0.$$

Since $k$ is arbitrary. This inequality implies that $f((b_n)_{n\in\mathbb{N}})=0$. Now, $(b_n)_{n\in\mathbb{N}}$ is any element of $B$. Thus, $f(b)=0$ for every $b\in B$.

Since $f\in V^*$ and $V=\text{span }A\cup B$ then $f=0$.