As pointed out in the comments already you made the mistake that you claimed the following: if for any $v \in V$ we have $F(v) =0$ then $v=0$. But this is only true if $F$ is injective, or equivalently if $\ker(F) = \{0\}$. So we cannot use this in this case.
Now we want to prove that given $a_1, \dots, a_n \in F$ (where $F$ is the underlying field of the vector spaces $V$ and $W$) we have that
$$a_1v_1+ \dots + a_nv_n=0 \Rightarrow a_1= \dots = a_n=0.$$
(this is the definition of linear independence of the vectors $v_1, \dots, v_n$).
In any reasonable exercise you should somehow use the assumption. So there has to be a point in our proof where we use that $F(v_1), \dots, F(v_n)$ are linearly independent. Let's start:
For the sake of contradiction assume that there is a non-trivial linear combination (non-trivial meaning that at least one coefficient $b_i$ is non-zero) of the $v_i$ that gives zero, i.e.
$$b_1v_1+ \dots b_nv_n=0 \text{ and at least one } b_i \ne 0.$$
(We now want to arrive at a contradiction because this assumption would mean that $v_1, \dots, v_n$ is linearly dependent).
Now from basic facts on linear operators we know that $F(0)=0$, therefore
$$0=F(0) = F(b_1v_1 + \dots b_nv_n) = b_1F(v_1) + \dots b_nF(v_n)$$
by properties 1. and 2. listed by you.
But if we compare the left hand side and the right hand side we have a linear combination of the vectors $F(v_1), \dots, F(v_n)$ giving zero, namely
$$0=b_1F(v_1) + \dots b_nF(v_n).$$
But remember: at least one $b_i$ is non-zero, so there is a non-trivial linear combination of $F(v_1), \dots, F(v_n)$ which gives zero. This is a contradiction of the assumption that $F(v_1), \dots, F(v_n)$ are linearly independent.
Hence our initial assumption must have been wrong and $v_1, \dots, v_n$ are linearly independent.
Best Answer
In questions like this, if you're still new to doing mathematics in general, I think it's really important to be really systematic:
In your case, $\{v_i\}_{1\leq i\leq n}$ is a basis for $V$. This means that $a)$, $\{v_i\}_{1\leq i\leq n}$ is a spanning set and $b)$, $\{v_i\}_{1\leq i\leq n}$ is a linearly independent.
So... given some $v\in V$, we need to show that it has a unique representation as a linear combination of $v_i$'s. So first, we need to show that it has such a representation. This follows from the fact that $\{v_i\}_{1\leq i\leq n}$ is spanning - this is literally just the definition of being spanning, so here, it is alright to say that it's obvious.
Next, we need to show uniqueness, and as you indicated, this should follow from linear independence. However, in order to do this, we need to reduce the question to a question of linear independence.
So, assume that $v=\sum_{i=1}^n \alpha_iv_i=\sum_{i=1}^n \beta_i v_i$. Then, we see that $0=\sum_{i=1}^n (\alpha_i-\beta_i)v_i$ and by definition of linear independence, this means that $\alpha_i-\beta_i=0$ for all $i$. This implies the desired.
The point of this type of exercise is typically to practice proof techniques, and the important thing, in my experience, is just to attempt to be completely systematic and never claim that something is just obvious (even though it may very well be).