So my teacher says that my proof for the theorem above is wrong. Would you mind pointing out the problems?
Let $A$ be an $n \times n$ matrix. If $A$ is diagonalizable, then every non-zero vector in $\mathbb R^n$ is an eigenvector.
Proof:
$A$ is diagonalizable so there is a basis $v_1,\ldots,v_n$ of $\mathbb R^n$ consisting of eigenvectors of $A$. There is therefore a scalar $\lambda$ such that $Av_i =\lambda v_i$. Let $v$ be an arbitrary vector in $\mathbb{R}^n$. Since $v_1,\ldots,v_n$ is a basis of $\mathbb R^n$, there must exist scalars $c_1,\ldots,c_n$ such that $v= c_1 v_1+\cdots+c_n v_n$.
$$\begin{align} Av &= A(c_1 v_1+\cdots+c_n v_n) \tag{1}\\ &= c_1 A v_1 +\cdots+ c_n A v_n \tag{2}\\ &= c_1(\lambda v_1)+\cdots+c_n(\lambda v_n) \tag{3}\\ &= \lambda (c_1 v_1+\cdots+c_n v_n) \tag{4} \\ & = \lambda v \tag{5} \end{align}$$
Therefore, $v$ is an eigenvector of $A$. $\square$