Pick a vector $v \in V \setminus \{ 0 \}$. You can find $v_2, \dots, v_n \in V$ such that $\{v,v_2, \dots,v_n \}$ is basis of $V$. Now consider the $(n-1)$-subspaces $W_i = \langle v, v_2, \dots, v_{i-1}, v_{i+1}, \dots, v_n \rangle$, for $i=2,\dots,n$. It is clear that the subspace $\langle v \rangle = W_2 \cap \cdots \cap W_n$ is $T$-invariant. So there exists $\lambda_v \in F$ such that $Tv = \lambda_v v$. (Pay attention: $\lambda_v$ depends on $v$.)
Now we must prove the scalars $\{ \lambda_v \}_{v \in V \setminus \{ 0 \}}$ are all the same. Choose two vectors v, v' \neq 0; if $v$ and v' are linearly dependent then it is clear that \lambda_v = \lambda_{v'}. If $v$ and v' are linearly independent, then \lambda_{v+v'} (v + v') = Tv + Tv' = \lambda_v v + \lambda_{v'} v', so \lambda_v = \lambda_{v+v'} = \lambda_{v'}.