How can I prove that if I have $n$ eigenvectors from different eigenvalues, they are all linearly independent?
I'll do it with two vectors. I'll leave it to you do it in general.

Suppose $\mathbf{v}_1$ and $\mathbf{v}_2$ correspond to distinct eigenvalues $\lambda_1$ and $\lambda_2$, respectively. 

Take a linear combination that is equal to $0$, $\alpha_1\mathbf{v}_1+\alpha_2\mathbf{v}_2 = \mathbf{0}$. We need to show that $\alpha_1=\alpha_2=0$.

Applying $T$ to both sides, we get
$$\mathbf{0} = T(\mathbf{0}) = T(\alpha_1\mathbf{v}_1+\alpha_2\mathbf{v}_2) = \alpha_1\lambda_1\mathbf{v}_1 + \alpha_2\lambda_2\mathbf{v}_2.$$
Now, instead, multiply the original equation by $\lambda_1$:
$$\mathbf{0} = \lambda_1\alpha_1\mathbf{v}_1 + \lambda_1\alpha_2\mathbf{v}_2.$$
Now take the two equations,
$$\begin{align*}
\mathbf{0} &= \alpha_1\lambda_1\mathbf{v}_1 + \alpha_2\lambda_2\mathbf{v}_2\\
\mathbf{0} &= \alpha_1\lambda_1\mathbf{v}_1 + \alpha_2\lambda_1\mathbf{v}_2
\end{align*}$$
and taking the difference, we get:
$$\mathbf{0} = 0\mathbf{v}_1 + \alpha_2(\lambda_2-\lambda_1)\mathbf{v}_2 = \alpha_2(\lambda_2-\lambda_1)\mathbf{v}_2.$$

Since $\lambda_2-\lambda_1\neq 0$, and since $\mathbf{v}_2\neq\mathbf{0}$ (because $\mathbf{v}_2$ is an eigenvector), then $\alpha_2=0$. Using this on the original linear combination $\mathbf{0} = \alpha_1\mathbf{v}_1 + \alpha_2\mathbf{v}_2$, we conclude that $\alpha_1=0$ as well (since $\mathbf{v}_1\neq\mathbf{0}$).

So $\mathbf{v}_1$ and $\mathbf{v}_2$ are linearly independent.

Now try using induction on $n$ for the general case.
