In this proof we will assume that $m=3$. We assume:
To prove that $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent we have to prove the following implication
\[
x_1 \mathbf{v}_1 + x_2 \mathbf{v}_2+ x_3 \mathbf{v}_3 = \mathbf{0} \quad \Rightarrow \quad x_1 = x_2 = x_3 = 0.
\]
To prove the preceding implication we assume
\begin{equation} \label{eq1}
x_1 \mathbf{v}_1 + x_2 \mathbf{v}_2+ x_3 \mathbf{v}_3 = \mathbf{0}.
\end{equation}
Step 1. Apply the matrix $A - \lambda_1 I$ to both sides of \eqref{eq1} to get
\begin{equation*}
(A - \lambda_1 I)\bigl(x_1 \mathbf{v}_1 + x_2 \mathbf{v}_2+ x_3 \mathbf{v}_3\bigr) = (A - \lambda_1 I)\mathbf{0}.
\end{equation*}
Next, we remember that this is a
linear algebra class, so we use the linearity property of $A - \lambda_1 I$ to get
\begin{equation} \label{eq1a}
x_1 (A - \lambda_1 I)\mathbf{v}_1 + x_2 (A - \lambda_1 I)\mathbf{v}_2+ x_3 (A - \lambda_1 I)\mathbf{v}_3 = \mathbf{0}.
\end{equation}
From
Assumption 3 it follows that
\begin{equation*}
(A - \lambda_1 I) \mathbf{v}_1 = \mathbf{0}, \quad
(A - \lambda_1 I) \mathbf{v}_2 = (\lambda_2 - \lambda_1) \mathbf{v}_2, \quad
(A - \lambda_1 I) \mathbf{v}_3 = (\lambda_3 - \lambda_1) \mathbf{v}_3.
\end{equation*}
Next we use the preceding three equalities in \eqref{eq1a} to obtain
\begin{equation} \label{eq2}
x_2 (\lambda_2 - \lambda_1) \mathbf{v}_2+ x_3 (\lambda_3 - \lambda_1) \mathbf{v}_3 = \mathbf{0}.
\end{equation}
Comment: The beauty of \eqref{eq2} is that the vector $\mathbf{v}_1$ has disappeared. In the next step we will disappear $\mathbf{v}_2.$
Step 2. Apply the matrix $A - \lambda_2 I$ to both sides of \eqref{eq2} and use
linear algebra to get
\begin{equation} \label{eq2a}
x_2 (\lambda_2 - \lambda_1) (A - \lambda_2 I) \mathbf{v}_2+ x_3 (\lambda_3 - \lambda_1) (A - \lambda_2 I) \mathbf{v}_3 = \mathbf{0}.
\end{equation}
From
Assumption 3 it follows that
\begin{equation*}
(A - \lambda_2 I) \mathbf{v}_2 = \mathbf{0}, \qquad
(A - \lambda_2 I) \mathbf{v}_3= (\lambda_3 - \lambda_2) \mathbf{v}_3.
\end{equation*}
Next we use the preceding two equalities in \eqref{eq2a} to obtain
\begin{equation} \label{eq3}
x_3 (\lambda_3 - \lambda_1) (\lambda_3 - \lambda_2) \mathbf{v}_3 = \mathbf{0}.
\end{equation}
Step 3.
By
Assumption 2 we have
\begin{equation*}
(\lambda_3 - \lambda_1) (\lambda_3 - \lambda_2) \neq 0.
\end{equation*}
Dividing \eqref{eq3} by $(\lambda_3 - \lambda_1) (\lambda_3 - \lambda_2)$ we obtain
\begin{equation*}
x_3 \mathbf{v}_3 = \mathbf{0}.
\end{equation*}
Since $\mathbf{v}_3 \neq \mathbf{0},$ we deduce that $x_3 = 0.$
Step 4. Substituting $x_3 = 0$ in \eqref{eq2} we get
\begin{equation} \label{eq4}
x_2 (\lambda_2 - \lambda_1) \mathbf{v}_2 = \mathbf{0}.
\end{equation}
Since by Assumption 2 we have $(\lambda_2 - \lambda_1) \neq 0,$ it follows from \eqref{eq4} that $x_2 \mathbf{v}_2 = \mathbf{0}.$ Since $\mathbf{v}_2 \neq \mathbf{0},$
we deduce that $x_2 = 0.$ Substituting $x_2 = x_3 = 0$ in \eqref{eq1} we get
$x_1 \mathbf{v}_1 = \mathbf{0}.$ Since $\mathbf{v}_1 \neq \mathbf{0},$
we deduce that $x_1 = 0.$ Thus we have proved that
\[
x_1 = 0, \qquad x_2 = 0, \qquad x_3 = 0.
\]
This completes the proof.
QED. The end of the proof.
Reasoning presented in the above proof is sometimes called recursive reasoning. We identify a certain process that leads to a desirible result (in this case eliminates a vector from a homogeneous vector equation), then we repeat that process many times to achieve a desired result. Sometimes this process is formally presented under the name Mathematical Induction.