I think this is true, but after searching can't find the confirmation I would expect. So I would appreciate confirmation or correction.
Proof:
A non-trivial vector space over $\mathbb R$ or $\mathbb C$ contains at least two different vectors, $0$ and $v \ne 0$ and by algebraic closure contains all linear combinations.
For any vector $a$ in an open set $A$ there is therefore $b \in V$ with $b \ne a$ and all $w_x = a + x.b \in V$ where $x \in R \ge 0$.
Since $A$ is open then there is some $r$ such that $ a \in B_r (a) \subset A$
$\|a - w_x\| = \|-x\cdot b\| = x\cdot\|b\|$ and for any $s > 0$ there is some $x(s)$ such that $x\cdot\|b\| < s$ and therefore $w_{x(s)} \in B_s(a)$.
So, for any $s$ with $r > s > 0$ then there is $w_{x(s)}$ with $a \ne w_{x(s)} \in A$ and $w_{x(s)} \in B_s(a) \subset B_r (a) \subset A$.
If $C$ is any open set containing $a$ then there is $t$ with $ a \in B_t (a) \subset C$. So for $s < min(r, t)$ then $w_x(s) \in A$ and $\in C$, i.e. every open set that contains $a$ contains a point of $A$ different to $a$ and so $a$ is a limit point of $A$.
Extension:
In fact $C$ contains all the points $w_x$ with $0 < x \le x(s)$ and so $a$ is a "condensation point" of $A$
I think your argument is right, but I find it too complicated. Here is a simpler argument. The key (trivial) fact is that if $t>0$ and $v\ne0$, then $tv\ne0$.
Now take $A$ open, $a\in A$. Fix any nonzero vector $v$. Then, as you say, there exists $r>0$ such that $B_r(a)\subset A$. $$ \|(a+tv)-a\|=\|tv\|=t\|v\|, $$ so $$\lim_{t\to0}a+tv=a. $$ When $t<r/\|v\|$, we have $a+tv\in B_r(a)\subset A$, so $a+tv\in A$ for all $t\in[0,r/\|v\|)$. And $a+tv\ne a$ if $t\ne0$.