Q1: If you know that $(v_j)_1^n$ are independent, then the direct sum decomposition holds naturally, because now the expression of $0$ as a sum of vectors from $E_j$ would be unique, then by definition the sum is a direct sum. If you want to prove the decomposition from the square one, you might use my answer as a reference.
Q2: I could give a proof.
We assume that $\{\lambda_j\}_1^n$ are distinct eigenvalues of a linear operator $\mathcal T \in \mathcal L(V)$.
Proof.$\blacktriangleleft$ Suppose $v_j \in E_j$ satisfy that
$v_1 + v_2 + \cdots + v_n =0$. By definition, $\mathcal T - \lambda_j \mathcal I$ is zero mapping on $E_j$. Therefore apply $\mathcal T - \lambda_1 \mathcal I$ to $\sum_1^n v_j = 0$ yields
$$
(\lambda_2 - \lambda_1) v_2 + (\lambda_3 - \lambda_1) v_3 + \cdots + (\lambda_n - \lambda_1) v_n = 0.
$$
Now apply $\mathcal T - \lambda_2 \mathcal I$ to it and obtain
$$
\sum_3^n (\lambda_j - \lambda_2) (\lambda_ j - \lambda_1) v_j = 0.
$$
Repeatedly we could know that if we apply
$$
(\mathcal T - \lambda_{n-1} \mathcal I)(\mathcal T -\lambda_{n-2} \mathcal I) \cdots (\mathcal T - \lambda_1 \mathcal I)
$$
to
$$
v_1 + v_2 + \cdots + v_n = 0,
$$
then we obtain
$$
\prod_{j=1}^{n-1} (\lambda_n - \lambda_j) v_n = 0.
$$
Since all $\lambda_j$ are distinct, $v_n = 0$.
Similarly, apply
$$
\prod_{j \neq k}(\mathcal T -\lambda_j\mathcal I) \quad [k = 1,2, \ldots, n-1]
$$
to $v_1 + \cdots + v_n =0 $ would yield similar expression
$$
\prod_{j \neq k} (\lambda_k - \lambda_j) v_k = 0,
$$
hence $v_k = 0$.
Conclusively, $v_j = 0$ for all $j$, as we desired. $\blacktriangleright$