Eigenspaces sum to the whole space

Induction on $r$. For $r=1$ there is nothing to prove. Suppose the result holds for $r-1$ eigenvalues. Let $v_i\in E_{\lambda_i}$ with $\sum_{i=1}^r v_i=0$. Applying $A$:
$$\sum_{i=1}^r \lambda_i v_i=0.$$
Subtracting $\lambda_r$ times the first relation:
$$\sum_{i=1}^{r-1}(\lambda_i-\lambda_r)v_i=0.$$
By the induction hypothesis applied to $\lambda_1,\ldots,\lambda_{r-1}$, each $(\lambda_i-\lambda_r)v_i=0$. Since $\lambda_i\neq\lambda_r$, $v_i=0$ for $i<r$. Then $v_r=0$ as well.

Conclusion. The eigenspaces are in direct sum, and $\sum_i\dim E_{\lambda_i}\le n$.