My attempt: $\exists c_1,\cdots,c_k\in F,~\exists v_1,\cdots,v_k\in V,~c_1v_1+\cdots+c_kv_k=0$, where $c_i$ are not all zero. Then $c_1T(v_1)+\cdots+c_kT(v_k)=0$. However, we can not deduce that $T(v_1),\cdots,T(v_k)$ are linearly dependent, since there may be $i,j$ such that $T(v_i)=T(v_j)$, so the coefficients $c_i$ and $c_j$ have the chance to sum up to zero!
For example, even though we can get that $2T(v_1)+0T(v_2)-2T(v_3)=0$, we cannot say that $T(v_1),T(v_2),T(v_3)$ is linearly dependent, since $T(v_1)$ and $T(v_3)$ could probably be the same!
Best Answer
You are confused with the definition of linear dependence. A $k$-tuple of vectors (like the $T(v_i)$) is linearly dependent as soon as there exists some non-trivial linear combination of them that gives zero. You've got such a combination here, the "non-trivial" is true because the $c_i$ were assumed from the outset to not all be zero. End of story.
To cut short on the comments, the statement as in your title, about linearly dependent sets is just not true. Mapping $T:\Bbb R^3\to\Bbb R^2$ by dropping the final coordinate is certainly linear, and the image under $T$ of the linearly dependent set $\{\,(1,0,z)\mid z\in\Bbb R\,\}$ is the linearly independent singleton set $\{(1,0)\}$. (Here $S$ is infinite, but taking any $3$ or more distinct vectors from it gives a linearly dependent finite set with the same property). The statement is true however when stated for $k$-tuples of vectors (linearly dependent $k$-tuple gives another linearly dependent $k$-tuple), and that is how it should have been stated.
The problem is that by laziness one often says "set" instead of "$k$-tuple" (and moreover, one much more often considers linearly independent vectors than dependent ones, in which case the laziness is relatively harmless, since being independent implies being distinct).