Prove there does not exist a linear map

The main problem with your proof is your understanding of linear dependence. Suppose that $$a_1v_1+\cdots+a_nv_n=0\ .$$ The statement that $v_1,\ldots,v_n$ are linearly dependent does not mean that the scalars are (definitely) not all zero, it means they are not necessarily all zero. Do you see the difference? To put it more formally, saying that the vectors are dependent means that there exist scalars, not all zero, for which the linear combination is zero.

Hint for a proof: if $v_1,\ldots,v_n$ are linearly dependent, then one of them can be written as a linear combination of the others, say $$v_1=a_1v_2+\cdots+a_nv_n\ .$$ Now the condition $$w_k=T(v_k)\quad\hbox{for all}\ k$$ (remember that you are trying to choose $w_k$ for which this is impossible) implies $$w_1=a_2w_2+\cdots+a_nw_n\ .$$ Can you choose $w_k$ such that this equation is definitely false?