Is the assumption $B \cap \operatorname{aint} (A)=\emptyset$ redundant in this version of Hahn-Banach theorem? I'm reading the proof of below theorem from this lecture note.

Theorem 0.10. Let $A$ and $B$ be two nonempty convex subsets of a vector space $X$. If $\operatorname{aint} (A) \neq \emptyset$ and $B \cap \operatorname{aint} (A)=\emptyset$, then $A, B$ can be separated by a hyperplane. Here $\operatorname{aint} (A)$ is the algebraic interior of $A$. If, moreover, $X$ is a t.v.s., $\operatorname{int}(A) \neq \emptyset$ and $B \cap \operatorname{int}(A)=\emptyset$, then $A$ and $B$ can be separated by a closed hyperplane.

The proof given by the author is as follows.

Proof. The set $A_{0} = \operatorname{aint} (A)$ is convex, nonempty and disjoint from $B$. Then the convex set $C=A_{0}-B$ has a nonempty algebraic interior and does no $0$. By Proposition 0.9., there exists a nontrivial linear functional $\ell$ on $X$ $\sup \ell(C) \leq \ell(0)=0$. It follows that
$$
\ell(a) \leq \ell(b) \quad \text { whenever } a \in A_{0}, b \in B .
$$
Thus $\sup \ell\left(A_{0}\right) \leq \inf \ell(B)$. By Observation 0.7., $\sup \ell(A) \leq \inf \ell(B)$. The proof of the second part proceeds in the same way, with $\operatorname{int}$ in place of $\operatorname{aint}$.
Remark 0.11. Let $A, B$ be as in the previous theorem. If a nontrivial linear functional $\ell$ separates $A$ and $B$, then it "separates strictly" $\operatorname{aint}(A)$ (or $\operatorname{int}(A)$) and $B$ in the following sense:
$$
\ell(a)<\inf \ell(B) \quad \text { whenever } a \in \operatorname{aint}(A) \quad (\text {or } a \in \operatorname{int}(A), \text { respectively })
$$
This follows immediately from Observation 0.7.

Related results are

*

*Proposition 0.9. Let $C$ be a convex subset of a vector space $X$. If $\operatorname{aint}(C) \neq \emptyset$ and $x_{0} \in X \setminus \operatorname{aint} (C)$, then $\left\{x_{0}\right\}$ and $C$ can be separated by a hyperplane. If, moreover, $X$ is a t.v.s. and $\operatorname{int}(C) \neq \emptyset$, then $\left\{x_{0}\right\}$ and $C$ can be separated by a closed hyperplane.


*Observation 0.7. Let $A$ be a convex subset of a vector space $X$ such that $\operatorname{aint} (A) \neq \emptyset$. Let $\alpha \in \mathbb R$ and $\ell: X \to \mathbb R$ be a nontrivial linear functional. Then the following statements are equivalent.



*

*(i) $A \subset \{\ell \le \alpha\}$.

*(ii) $\operatorname{aint} (A) \subset \{\ell \le \alpha\}$.

*(iii) $\operatorname{aint} (A) \subset \{\ell < \alpha\}$.


In my below attempt, I only use Proposition 0.9. and don't use the assumption that $B \cap \operatorname{aint} (A)=\emptyset$. Could you confirm if my proof is fine?
My attempt: WLOG, we assume $0 \in B$. Then $A \subset C := A-B$. Also, $C$ is convex and $0 \notin C$. Let $a \in \operatorname{aint} (A)$. For $x \in X$, there is $t>0$ such that $[a, a+tx] \subset A \subset C$. It follows that $a \in \operatorname{aint} (C)$. This means $\operatorname{aint} (A) \subset \operatorname{aint} (C) \neq \emptyset$. By our Proposition 0.9., there is a linear functional $\ell:X \to \mathbb R$ such that $\sup \ell(C) \le \ell(0)=0$, i.e.,
$$
\ell(a) \le \ell(b) \quad \forall a\in A, b \in B. 
$$
Hence $\sup \ell(A) \le \inf \ell (B)$. For the last part, notice that if $X$ is a t.v.s. and $\operatorname{int}(A) \neq \emptyset$, then $\operatorname{int} (C) \neq \emptyset$.
 A: As @Ruy mentioned in a comment, we can not get rid of the assumption $B \cap \operatorname{aint} (A)=\emptyset$. I actually misunderstood. In my attempt, I claim that $0 \notin C := A-B$. By this claim, I assume $A\cap B = \emptyset$, which is actually stronger than $B \cap \operatorname{aint} (A)=\emptyset$. Below is my corrected proof.

Let $A' := \operatorname{aint} (A)$. Because $A$ is convex, we get $A'$ is convex and $\operatorname{aint} (A') = A'$. WLOG, we assume $0 \in B$. Then $A' \subset C := A' - B$. We have $C$ is convex and $0 \notin C$. Let $a \in A'$. Then $a \in \operatorname{aint} (A')$, i.e., for $x \in X$, there is $t>0$ such that $[a, a+tx] \subset A' \subset C$. It follows that $a \in \operatorname{aint} (C)$. This means $A' \subset \operatorname{aint} (C) \neq \emptyset$. By Proposition 0.9., there is a linear functional $\ell:X \to \mathbb R$ such that $\sup \ell(C) \le \ell(0)=0$, i.e.,
$$
\ell(a) \le \ell(b) \quad \forall a\in A', b \in B. 
$$
Hence $\sup \ell(A') \le \inf \ell (B)$. By Observation 0.7., $\sup \ell(A) \leq \inf \ell(B)$.
For the last part, if $X$ is a t.v.s. and $\operatorname{int}(A) \neq \emptyset$, then $\operatorname{int} (C) \neq \emptyset$. Indeed, pick $a \in \operatorname{int}(A) \subset A' \subset C$. Because $\operatorname{int}(A)$ is open, there is a neighborhood $V$ of $a$ such that $V \subset \operatorname{int}(A) \subset C$. It follows that $a \in \operatorname{int}(C)$.
