I think it is a True statement, and my proof is as follows:

let $\theta $ be the additive identity in F

Given any element v in V, since v + v = $\theta$ v + $\theta$ v = 2$\theta$ v, then 2$\theta$ $\in$ F, and $\frac{\theta}{2} \in$ F, hence $\frac{\theta}{2}v \in$ V, by this argument we can construct infinite $\{2v, \frac{\theta}{2}v , 3v, \frac{\theta}{3}v...\}$

Hence V cannot be finite, and we are done.