Let $V$ be a $\Bbb{R}$-vector space. In the paper I am reading, the author wants to
"...equip $V$ with a locally convex topology, called the 'algebraic topology' (also known as the finest locally convex topology), by declaring that any convex set $C$ such that $int(C) = C$ is open."
Why is this a topology? The union of two convex sets is not necessarily convex, so the collection might not be closed under arbitrary unions. I thought that, perhaps, the author was claiming that the collection formed a basis, but I don't see how it forms a basis. By the way, $int(C)$ denotes the set of 'algebraic' interior points; ie., $c \in C$ is an algebraic interior point if for every $v \in V$, there is a $t \in (0,1]$ such that $(1-t)c + tv \in C$.
Using convex sets is probably not the best way to explain the topology. Given a vector space $V$ and a set $A$ of seminorms on $V$, one can define a locally convex topology on $V$. The precise definition is explained here:
Doubt in understanding Space $D(\Omega)$
A particular case of this construction is when one takes for $A$ the set of all seminorms on $V$ (denoted by $s(V)$ in the above linked post). This is how you get the finest locally convex topology. It is "algebraic" in the sense that the only input needed is the algebraic structure of $V$ as a vector space on $\mathbb{R}$ or $\mathbb{C}$. For some $V$'s this is also the most appropriate topology to use for that space. Examples of that are the space of almost finite sequences (or polynomials), as well as the Schwartz-Bruhat space over the $p$-adics.
Finally, the relation to convex sets is via the Minkowski Gauge Theorem, but I think it is an unnecessary and perhaps pedagogically counterproductive detour.