I'm trying to prove the second statement of below result, taken from this page of Wikipedia.
In my below proof, I assume $X$ to be Hausdorff. The Hausdorff condition is to ensure that every finite-dimensional Hausdorff t.v.s. is topologically isomorphic to some $\mathbb R^d$. This result is then used in the proof of Lemma 1.
Could you have a check on my attempt?
Does the theorem hold in case $X$ is non-Hausdorff?
Theorem: Let $A$ be a convex subset of a Hausdorff t.v.s. $X$. If $\dim A < +\infty$ then $\operatorname{aint} (A) = \operatorname{int} (A)$.
Lemma 1: Let $V$ be a $n$-dimensional Hausdorff t.v.s. and $A:=\{x_1, \ldots, x_{n+1}\} \subset V$ affinely independent. Then $\operatorname{int} (\operatorname{conv} A) \neq \emptyset$.
Lemma 2: Let $A$ be a convex subset of a t.v.s. $X$. If $\operatorname{int} (A) \neq \emptyset$, then $\operatorname{aint} (A) = \operatorname{int} (A)$.
Proof: If $\operatorname{aint} (A) = \emptyset$, then we are done. Now consider the case $\operatorname{aint} (A) \neq \emptyset$. WLOG, we assume $0 \in \operatorname{aint} (A)$. For each $x \in X$, there is $t>0$ such that $tx \in A$. Then $A$ contains a basis of $X$, so $X = \operatorname{aff} (A)$. Hence $X$ is finite-dimensional. By Lemma 1, we get $\operatorname{int} (\operatorname{conv} A) \neq \emptyset$. Because $A$ is convex, $\operatorname{conv} A = A$ and thus $\operatorname{int} A \neq \emptyset$. The claim then follows from Lemma 2.
