This is an extended version of my observations in the comments.
(1) There are pointed convex cones that do not have a base. To see this, take $V=\mathbb{R}^2$ as a simple example, with $C$ given by all those $(x,y)\in\mathbb{R}^2$ for which $x>0$, or $x=0$ and $y>0$. One can visualize the non-existence of a convex base by intersecting $C$ with a line not passing through the origin, and then rotating this line all the way around, noticing that it never hits all the rays of $C$.
More formally, suppose that $K$ is a convex base. This means that there is exactly one point of the form $(0,y_0)\in K$ with $y_0\in K$; up to rescaling, we may assume $y_0=1$. Similarly, we can assume that $(1,0)\in K$ is the only element in $K$ on the $x$-axis.
But then also the ray generated by $(1,-1)$ must intersect $K$ somewhere, say at $(t,-t)$ for $t>0$. By convexity and $(0,1)\in K$, this would imply that the point $$ \frac{t}{t+1}(0,1) + \frac{1}{t+1}(t,-t) = \left(\frac{t}{t+1},0\right) $$ is in $K$ as well, in contradiction to the assumption that $(1,0)$ is the only point in $K$ on the $x$-axis.
(2) Every convex cone has a base (not necessarily convex). This is a simple consequence of a axiom of choice: we can simply pick one representative of each ray $(C\setminus\{0\})/(\mathbb{R}_{>0})$.
In conclusion, there exist cones all of whose bases are non-convex. Therefore what the OP is trying to do is bound not to work.
Arguably, the non-existence of a convex base in (1) is due to the cone not being closed. In fact, in finite dimensions, one can show that every pointed closed convex cone has a base, using the Hahn-Banach theorem as suggested by Willie Wong in the comments. However, since constructing a convex base like this requires a functional which is strictly positive on the whole cone. In fact, a pointed closed cone in locally convex spaces does not necessarily have a convex base either: see Exercise 1.7.6 in Aliprantis/Tourky, "Cones and Duality".